Question 15 Marks
Explain why,
- A horse cannot pull a cart and run in empty space.
- Passengers are thrown forward from their seats when a speeding bus stops suddenly.
- It is easier to pull a lawn mower than to push it.
- A cricketer moves his hands backwards while holding a catch.
Answer
- While trying to pull a cart, ahorse pushes the ground backward with some force. The ground in turn exerts an equal and opposite reaction force upon the feet of the horse. This reaction force causes the horse to move forward.
An empty space is devoid of any such reaction force. Therefore, a horse cannot pull a cart and run in empty space.
- This is due to inertia of motion. When a speeding bus stops suddenly, the lower part of a passenger’s body, which is in contact with the seat, suddenly comes to rest. However, the upper part tends to remain in motion (as per the first law of motion). As a result, the passenger’s upper body is thrown forward in the direction in which the bus was moving.
- While pulling a lawn mower, a force at an angle $\theta$ is applied on it, as shown in the following figure.
The vertical component of this applied force acts upward. This reduces the effective weight of the mower. On the other hand, while pushing a lawn mower, a force at an angle $\theta$ is applied on it, as shown in the following figure.
In this case, the vertical component of the applied force acts in the direction of the weight of the mower. This increases the effective weight of the mower.
Since the effective weight of the lawn mower is lesser in the first case, pulling the lawn mower is easier than pushing it.
- According to Newton’s second law of motion, we have the equation of motion:
$\text{F}=\text{ma}=\frac{\text{m}\triangle\text{v}}{\triangle\text{t}}\ ....(\text{i})$
Where,
F = Stopping force experienced by the cricketer as he catches the ball
m = Mass of the ball
$\triangle\text{t}$ = Time of impact of the ball with the hand
It can be inferred from equation (i) that the impact force is inversely proportional to the impact time, i.e.,
$\frac{\text{F}\propto1}{\triangle\text{t}}\ ...(\text{ii})$
Equation (ii) shows that the force experienced by the cricketer decreases if the time of impact increases and vice versa.
While taking a catch, a cricketer moves his hand backward so as to increase the time of impact $(\triangle\text{t})$. This is turn results in the decrease in the stopping force, thereby preventing the hands of the cricketer from getting hurt. View full question & answer→Question 25 Marks
A truck starts from rest and accelerates uniformly at $2.0ms^{-2}$. At $t = 10s$, a stone is dropped by a person standing on the top of the truck (6m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at $t = 11s$? (Neglect air resistance.)
Answer
- Initial velocity of the truck, u = 0
Acceleration, $a = 2m/s^2$
Time, t = 10s
As per the first equation of motion, final velocity is given as:
v = u + at
= 0 + 2 × 10 = 20m/s
The final velocity of the truck and hence, of the stone is 20m/s.
At t = 11s, the horizontal component $(v_x)$ of velocity, in the absence of air resistance, remains unchanged, i.e.,
vx = 20m/s
The vertical component (v_y) of velocity of the stone is given by the first equation of motion as:
$\text{v}_\text{y}=\text{u}+\text{ay}\delta\text{t}$
Where, $\delta\text{t} = 11 - 10 = 1s$ and $ay = g = 10m/s^2$
$\therefore v_y = 0 + 10 \times 1 = 10m/s$
The resultant velocity (v) of the stone is given as:
$v = (v_x^2 + v_y^2)^{1/2}$
$= (20^2 + 10^2)^{1/2}$
$= 22.36m/s$
Let $\theta$ be the angle made by the resultant velocity with the horizontal component of velocity, $v_x$
$\therefore\tan\theta=\frac{\text{v}_\text{y}}{\text{v}_\text{x}}$
$\theta=\tan^{-1}\frac{10}{20}$
$=26.57^\circ$
- When the stone is dropped from the truck, the horizontal force acting on it becomes zero. However, the stone continues to move under the influence of gravity. Hence, the acceleration of the stone is $10 m/s^2$ and it acts vertically downward.
View full question & answer→Question 35 Marks
A helicopter of mass $1000kg$ rises with a vertical acceleration of $15ms^{-2}$. The crew and the passengers weigh $300kg$. Give the magnitude and direction of the,
- Force on the floor by the crew and passengers.
- Action of the rotor of the helicopter on the surrounding air.
- Force on the helicopter due to the surrounding air.
AnswerMass of the helicopter, $m_h = 1000kg$ Mass of the crew and passengers, $m_p = 300kg$ Total mass of the system, $m = 1300kg$ Acceleration of the helicopter, $a = 15m/s^2$
- Using Newton’s second law of motion, the reaction force R, on the system by the floor can be calculated as:
$R - m_pg = ma$
$= m_p(g + a)$
$= 300(10 + 15) = 300 \times 25$
$= 7500N$
Since the helicopter is accelerating vertically upward, the reaction force will also be directed upward. Therefore, as per Newton’s third law of motion, the force on the floor by the crew and passengers is 7500N, directed downward.
Using Newton’s second law of motion, the reaction force R’, experienced by the helicopter can be calculated as:
R' - mg = ma
= m(g + a)
= 1300(10 + 15) = 1300 × 25
= 32500N
- The reaction force experienced by the helicopter from the surrounding air is acting upward. Hence, as per Newton’s third law of motion, the action of the rotor on the surrounding air will be 32500 N, directed downward.
- The force on the helicopter due to the surrounding air is 32500N, directed upward.
View full question & answer→Question 45 Marks
Two bodies A and B of masses $5kg$ and $10kg$ in contact with each other rest on a table against a rigid wall (Fig.). The coefficient of friction between the bodies and the table is $0.15$. A force of $200N$ is applied horizontally to A. What are (a) the reaction of the partition (b) the action-reaction forces between A and B? What happens when the wall is removed? Does the answer to (b) change, when the bodies are in motion? Ignore the difference between $µ_s$ and $µ_k$.
Answer
- Mass of body A, $m_A = 5kg$ Mass of body B, $m_B = 10kg$
Applied force, F = 200N
Coefficient of friction, $\mu\text{s}=0.15$
The force of friction is given by the relation:
$\text{f}_\text{s}=\mu(\text{m}_\text{A}+\text{m}_\text{B})\text{g}$
= 0.15(5 + 10) × 10
= 1.5 × 15 = 22.5N leftward
Net force acting on the partition = 200 - 22.5 = 177.5N rightward
As per Newton’s third law of motion, the reaction force of the partition will be in the direction opposite to the net applied force.
Hence, the reaction of the partition will be 177.5N, in the leftward direction.
- Force of friction on mass A:
$\text{f}_\text{A}=\mu\text{m}_\text{A}\text{g}$
= 0.15 × 5 × 10 = 7.5N leftward
Net force exerted by mass A on mass B = 200 - 7.5 = 192.5N rightward
As per Newton’s third law of motion, an equal amount of reaction force will be exerted by mass B on mass A, i.e., 192.5N acting leftward.
When the wall is removed, the two bodies will move in the direction of the applied force.
Net force acting on the moving system = 177.5N
The equation of motion for the system of acceleration a,can be written as:
Net force = $(m_A + m_B)a$
$\therefore\text{a}=\frac{\text{Net force}}{(\text{m}_\text{A}+\text{m}_\text{B})}$
$=\frac{177.5}{(5+10)}=\frac{177.5}{15}=11.83\text{ms}^{-2}$
Net force causing mass A to move:
$F_A = m_Aa$
= 5 × 11.83 = 59.15N
Net force exerted by mass A on mass B = 192.5 - 59.15 = 133.35N
This force will act in the direction of motion. As per Newton’s third law of motion, an equal amount of force will be exerted by mass B on mass A, i.e., 133.3N, acting opposite to the direction of motion. View full question & answer→Question 55 Marks
The rear side of a truck is open and a box of $40kg$ mass is placed 5m away from the open end as shown in Fig. The coefficient of friction between the box and the surface below it is $0.15$. On a straight road, the truck starts from rest and accelerates with $2ms^{-2}$. At what distance from the starting point does the box fall off the truck? (Ignore the size of the box).
AnswerMass of the box, m = 40kg Coefficient of friction, $\mu=0.15$ Initial velocity, u = 0 Acceleration, $a = 2m/s^2$ Distance of the box from the end of the truck, s' = 5m As per Newton’s second law of motion, the force on the box caused by the accelerated motion of the truck is given by: $F = ma = 40 \times 2 = 80N$ As per Newton’s third law of motion, a reaction force of 80N is acting on the box in the backward direction. The backward motion of the box is opposed by the force of friction f, acting between the box and the floor of the truck. This force is given by: $\text{f}=\mu\text{mg} = 0.15 \times 40 \times 10 = 60N$
$\therefore$ Net force acting on the block: $F_{net} = 80 - 60 = 20N$ backward The backward acceleration produced in the box is given by: $\text{a}_\text{back}=\frac{\text{F}_\text{net}}{\text{m}}=\frac{20}{40}=0.5\text{ms}^{-2}$ Using the second equation of motion, time t can be calculated as: $\text{s}'=\text{ut}+\Big(\frac{1}{2}\Big)\text{a}_\text{back}\text{t}^2$
$5=0+\Big(\frac{1}{2}\Big)\times0.5\times\text{t}^2$
$\therefore\text{t}=\sqrt{20}\text{s}$ Hence, the box will fall from the truck after $\sqrt{20}\text{s}$ from start. The distance s, travelled by the truck in $\sqrt{20}\text{s}$ is given by the relation: $\text{s}=\text{ut}+\Big(\frac{1}{2}\Big)\text{at}^2$
$=0+\Big(\frac{1}{2}\times2\times\Big(\sqrt{20}\Big)^2$
$=20\text{m}$
View full question & answer→Question 65 Marks
Two masses $8kg$ and $12kg$ are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.
AnswerThe given system of two masses and a pulley can be represented as shown in the following figure: 
Smaller mass, $\mathrm{m}_1=8 \mathrm{~kg}$ Larger mass, $\mathrm{m}_2=12 \mathrm{~kg}$ Tension in the string $=\mathrm{T}$ Mass $\mathrm{m}_2$, owing to its weight, moves downward with acceleration a, and mass $m_1$ moves upward. Applying Newton's second law of motion to the system of each mass: For mass $m_1$ : The equation of motion can be written as: $T-m_1 g=$ ma ...(i) For mass $m_2$ : The equation of motion can be written as: $\mathrm{m}_2 \mathrm{~g}-\mathrm{T}=\mathrm{m}_2 \mathrm{a}$...(ii) Adding equations (i) and (ii), we get: $\left(\mathrm{m}_2-\mathrm{m}_1\right) \mathrm{g}=\left(\mathrm{m}_1+\mathrm{m}_2\right) \mathrm{a}$ $\therefore \mathrm{a}=\frac{\left(\mathrm{m}_2-\mathrm{m}_1\right)}{\left(\mathrm{m}_2-\mathrm{m}_1\right) \mathrm{g}} \ldots$ (iii) $=\frac{12-8}{12+8} \times 10=4 \times \frac{10}{20}=2 \mathrm{~ms}^{-2}$ Therefore, the acceleration of the masses is $2 \mathrm{~m} / \mathrm{s}^2$. Substituting the value of a in equation (ii), we get, $m_2 g-T \frac{m_2\left(m_2-m_1\right)}{\left(m_1-m_2\right)} T=\frac{m_2-\left(m_2^2-m_1 m_2\right)}{\left(m_1+m_2\right) g}=\frac{2 m_1 m_2 g}{m_1+m_2}=\frac{2 \times 12 \times 8 \times 10}{12+8}$ $=96 \mathrm{~N}$ Therefore, the tension in the string is 96 N . View full question & answer→Question 75 Marks
Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m. Give the magnitude and direction of:
- The force on the $7^{th}$ coin (counted from the bottom) due to all the coins on its top.
- The force on the $7^{th}$ coin by the eighth coin.
- The reaction of the 6th coin on the $7^{th}$ coin.
Answer
- Force on the seventh coin is exerted by the weight of the three coins on its top.
Weight of one coin = mg.
Weight of three coins = 3mg
Hence, the force exerted on the $7^{th}$ coin by the three coins on its top is 3mg. This force acts vertically downward.
- Force on the seventh coin by the eighth coin is because of the weight of the eighth coin and the other two coins (ninth and tenth) on its top.
Weight of the eighth coin = mg
Weight of the ninth coin = mg
Weight of the tenth coin = mg
Total weight of these three coins = 3mg
Hence, the force exerted on the $7^{th}$ coin by the eighth coin is 3mg. This force acts vertically downward.
- The $6^{th}$ coin experiences a downward force because of the weight of the four coins ($7^{th}, 8^{th}, 9^{th},$ and $10^{th}$) on its top.
Therefore, the total downward force experienced by the $6^{th}$ coin is 4mg.
As per Newton's third law of motion, the 6th coin will produce an equal reaction force on the $7^{th}$ coin, but in the opposite direction. Hence, the reaction force of the $6^{th}$ coin on the $7^{th}$ coin is of magnitude 4mg. This force acts in the upward direction. View full question & answer→Question 85 Marks
Figure $5.17$ shows the position-time graph of a body of mass $0.04kg$. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the body? What is the magnitude of each impulse?
AnswerA ball rebounding between two walls located between at x = 0 and x = 2cm; after every 2s, the ball receives an impulse of magnitude $0.08 \times 10^{-2}kgm/s$ from the walls The given graph shows that a body changes its direction of motion after every 2s. Physically, this situation can be visualized as a ball rebounding to and fro between two stationary walls situated between positions x = 0 and x = 2cm. Since the slope of the x - t graph reverses after every 2s, the ball collides with a wall after every 2s. Therefore, ball receives an impulse after every 2s. Mass of the ball, m = 0.04kg The slope of the graph gives the velocity of the ball. Using the graph, we can calculate initial velocity (u) as: $\text{u}=\frac{(2-0)\times10^{-2}}{(2-0)}=10^{-2}\text{m/s}$ Velocity of the ball before collision, $u = 10^{-2}m/s$ Velocity of the ball after collision, $v = -10^{-2}m/s$ (Here, the negative sign arises as the ball reverses its direction of motion.) Magnitude of impulse = Change in momentum $= |mv - mu| = |0.04 (v - u)| = |0.04(-10^{-2} - 10^{-2})| = 0.08 \times 10^{-2}kgm/s$
View full question & answer→Question 95 Marks
Figure. shows the position-time graph of a particle of mass 4kg. What is the (a) force on the particle for t < 0, t > 4s, 0 < t < 4s? (b) impulse at t = 0 and t = 4s? (Consider one-dimensional motion only).
Answer
- For t < 0
It can be observed from the given graph that the position of the particle is coincident with the time axis. It indicates that the displacement of the particle in this time interval is zero. Hence, the force acting on the particle is zero.
For t > 4s
It can be observed from the given graph that the position of the particle is parallel to the time axis. It indicates that the particle is at rest at a distance of
3 m from the origin. Hence, no force is acting on the particle.
For 0 < t < 4
It can be observed that the given position-time graph has a constant slope. Hence, the acceleration produced in the particle is zero. Therefore, the force acting on the particle is zero.
- At t = 0
Impulse = Change in momentum
= (mv - mu)
Mass of the particle, m = 4kg
Initial velocity of the particle, u = 0
Final velocity of the particle $\text{v}=\frac{3}{4}\text{m/s}$
Therefore, Impulse $=4\Big(\frac{3}{4}-0\Big)=3\text{kg m/s}$
At t = 4s
Initial velocity of the particle, $\text{u}=\frac{3}{4}\text{m/s}$
Final velocity of the particle, v = 0
Therefore, Impulse $=4\Big(0-\frac{3}{4}\Big)=-3\text{kg m/s}$ View full question & answer→Question 105 Marks
A body of mass $0.40kg$ moving initially with a constant speed of $10ms^{-1}$ to the north is subject to a constant force of $8.0 N$ directed towards the south for 30s. Take the instant the force is applied to be $t = 0$, the position of the body at that time to be $x = 0$, and predict its position at $t = –5s, 25s, 100s$.
AnswerMass of the body, m = 0.40kg Initial speed of the body, u = 10m/s due north Force acting on the body, F = -8.0N Acceleration produced in the body, $\text{a}=\frac{\text{F}}{\text{m}}=\frac{-8.0}{0.40}=-20\text{ms}^{-2}$
- At t = -5s
Acceleration, a‘ = 0 and u = 10m/s
$\text{s}=\text{ut}+\frac{1}{2}\text{a' t}^2$
= 10 × (-5) = -50m
- At t = 25s
Acceleration, $a” = -20m/s^2$ and $u = 10m/s$
$\text{s}'=\text{ut}'+\frac{1}{2}\text{a'' t}^2$
$=10\times25+\frac{1}{2}\times-20\times(25)^2$
= 250 - 6250 = -6000m
- At t = 100s
For $0<\text{t}\le30\text{s}$
$a = -20ms^{-2}$
$u = 10m/s$
$\text{s}_1=\text{ut}+\frac{1}{2}\text{a}"\text{t}^2$
$=10\times30+\frac{1}{2}\times-20\times(30)^2$
= 300 - 9000 = -8700m
For $30<\text{t}\le100\text{s}$
As per the first equation of motion, for t = 30s, final velocity is given as:
v = u + at
= 10 + (-20) × 30 = -590m/s
Velocity of the body after 30s = -590m/s
For motion between 30s to 100s, i.e., in 70s:
$\text{s}_2=\text{vt}+\frac{1}{2}\text{a}"\text{t}^2$
= -590 × 70 = -41300m
$\therefore$ Total distance, $s” = s_1 + s_2 = -8700 - 41300 = -50000m = -50km$. View full question & answer→Question 115 Marks
A disc revolves with a speed of $33\frac{1}{3}\text{rev/min},$ and has a radius of $15cm$. Two coins are placed at $4cm$ and $14cm$ away from the centre of the record. If the co-efficient of friction between the coins and the record is $0.15$, which of the coins will revolve with the record?
AnswerCoin placed at 4cm from the centre Mass of each coin = m Radius of the disc, $\text{r}=15\text{m}=0.15\text{m}$ Frequency of revolution, $\text{v}=\frac{100}{3}\text{rev/min}=\frac{100}{3\times60}=\frac{5}{9}\text{rev/s}$ Coefficient of friction, $\mu=0.15$ In the given situation, the coin having a force of friction greater than or equal to the centripetal force provided by the rotation of the disc will revolve with the disc. If this is not the case, then the coin will slip from the disc. Coin placed at 4cm: Radius of revolution, $\text{r}'=4\text{cm}=0.04\text{m}$ Angular frequency, $\omega=2\pi\text{v}=2\times\frac{22}{7}\times\frac{5}{9}=3.49\text{s}^{-1}$ Frictional force, $\text{f}=\mu\text{mg}=0.15\times\text{m}=10=1.5\text{mN}$ Centripetal force on the coin: $\text{F}_\text{cent}=\text{mr},\omega^2$
$=\text{}\text{m}\times0.04\times(3.49)^2$
$=0.49\text{mN}$ Since $f > F_{cent}, the coin will revolve along with the record. Coin placed at 14cm: Radius, $\text{r"}=14\text{cm}=0.14\text{m}$ Angular frequency, $\omega=3.49\text{s}^{-1}$ Frictional force, $\text{f}'=1.5\text{mN}$ Centripetal force is given as: $\text{F}_\text{cent}=\text{mr},\omega^2$
$=\text{}\text{m}\times0.14\times(3.49)^2$
$=1.7\text{mN}$ Since $f < F_{cent}$, the coin will slip from the surface of the record.
View full question & answer→Question 125 Marks
A block of mass $15kg$ is placed on a long trolley. The coefficient of static friction between the block and the trolley is $0.18$. The trolley accelerates from rest with $0.5ms^{-2}$ for $20s$ and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground, (b) an observer moving with the trolley.
Answer
- Mass of the block, m = 15kg
Coefficient of static friction, $\mu=0.18$
Acceleration of the trolley, $a = 0.5m/s^2$
As per Newton's second law of motion, the force (F) on the block caused by the motion of the trolley is given by the relation:
F = ma = 15 × 0.5 = 7.5N
This force is acted in the direction of motion of the trolley.
Force of static friction between the block and the trolley:
$\text{f}=\mu\text{mg}=0.18\times15\times10=27\text{N}$
The force of static friction between the block and the trolley is greater than the applied external force. Hence, for an observer on the ground, the block will appear to be at rest.
When the trolley moves with uniform velocity there will be no applied external force. Only the force of friction will act on the block in this situation.
- An observer, moving with the trolley, has some acceleration. This is the case of non-inertial frame of reference. The frictional force, acting on the trolley backward, is opposed by a pseudo force of the same magnitude. However, this force acts in the opposite direction. Thus, the trolley will appear to be at rest for the observer moving with the trolley.
View full question & answer→Question 135 Marks
A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are: [Choose the correct alternative]
|
|
Lowest Point
|
Highest Point
|
|
(a)
|
$mg - T_1$
|
$mg + T_2$
|
|
(b)
|
$mg + T_1$
|
$mg - T_2$
|
|
(c)
|
$mg + T_1 - (mv_1^2)/R$
|
$mg - T_2 + (mv_1^2)/R$
|
|
(d)
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$mg - T_1 - (mv_1^2)/R$
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$mg + T_2 + (mv_1^2)/R$
|
$T_1$ and $v_1$ denote the tension and speed at the lowest point. $T_2$ and $v_2$ denote corresponding values at the highest point. AnswerThe free body diagram of the stone at the lowest point is shown in the figure below:
According to Newton’s second law of motion, the net force acting on the stone at this point is equal to the centripetal force. i.e., $F_{net} = T - mg \frac{\text{mv}_1^2}{\text{R}}\ ....(\text{i})$ where, $v_1$ is the velocity at the lowest point. The free body diagram of the stone at the highest point is shown in the following figure.
Using Newton’s second law of motion, $\text{T}+\text{mg}=\frac{\text{mv}^2_2}{\text{R}}\ ....(\text{ii})$ Where, $v_2$ is the velocity at the highest point. From equations (i) and (ii), Net force acting at the lowest = (T - mg) Net force at the highest points = (T + mg). View full question & answer→Question 145 Marks
A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg)
AnswerThe given situation can be represented as shown in the following figure. 
Where, AO = Incident path of the ball OB = Path followed by the ball after deflection $\angle\text{AOB}$ = Angle between the incident and deflected paths of the ball = 45° $\angle\text{AOP}=\angle\text{BOP}=22.5^\circ=\theta$ Initial and final velocities of the ball = v Horizontal component of the initial velocity = $\text{v}\cos\theta$ along RO Vertical component of the initial velocity = $\text{v}\sin\theta$ along PO Horizontal component of the final velocity = $\text{v}\cos\theta$ along OS Vertical component of the final velocity = $\text{v}\cos\theta$ along OP The horizontal components of velocities suffer no change. The vertical components of velocities are in the opposite directions. $\therefore$ Impulse imparted to the ball = Change in the linear momentum of the ball $=\text{mv}\cos\theta-(-\text{mv}\cos\theta)=2\text{mv}\cos\theta$ Mass of the ball, m = 0.15kg Velocity of the ball, v = 54km/h = 15m/s $\therefore$ Impulse $2\times0.15\times15\cos22.5^\circ=4.16\text{kgm/s}$ View full question & answer→Question 155 Marks
A stream of water flowing horizontally with a speed of $15ms^{-1}$ gushes out of a tube of cross-sectional area $10^{-2}m^2$, and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound?
AnswerSpeed of the water stream, $\mathrm{v}=15 \mathrm{~m} / \mathrm{s}$ Cross-sectional area of the tube, $\mathrm{A}=10^{-2} \mathrm{~m}^2$ Volume of water coming out from the pipe per second, $V=A v=15 \times 10^{-2} \mathrm{~m}^3 / \mathrm{s}$ Density of water, $p=10^3 \mathrm{~kg} / \mathrm{m}^3$ Mass of water flowing out through the pipe per second $=p \times V=150 \mathrm{~kg} / \mathrm{s}$ The water strikes the wall and does not rebound. Therefore, according to Newton's second law of motion, Force exerted by the water on the wall, F = Rate of change of momentum $=\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}=\frac{\mathrm{mv}}{\mathrm{t}} 150 \times 15=2250 \mathrm{~N}$
View full question & answer→Question 165 Marks
A block of mass 25kg is raised by a 50kg man in two different ways as shown in Fig. What is the action on the floor by the man in the two cases? If the floor yields to a normal force of 700N, which mode should the man adopt to lift the block without the floor yielding?
Answer750N and 250N in the respective cases; Method (b) Mass of the block, m = 25kg Mass of the man, M = 50kg Acceleration due to gravity, $g = 10m/s^2$ Force applied on the block, F = 25 × 10 = 250N Weight of the man, W = 50 × 10 = 500N Case (a): When the man lifts the block directly In this case, the man applies a force in the upward direction. This increases his apparent weight. $\therefore$ Action on the floor by the man = 250 + 500 = 750N Case (b): When the man lifts the block using a pulley In this case, the man applies a force in the downward direction. This decreases his apparent weight. $\therefore$ Action on the floor by the man = 500 - 250 = 250N If the floor can yield to a normal force of 700N, then the man should adopt the second method to easily lift the block by applying lesser force.
View full question & answer→Question 175 Marks
Two bodies of masses $10kg$ and $20kg$ respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force $F = 600N$ is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case?
AnswerHorizontal force, $F=600 \mathrm{~N}$ Mass of body $A, m_1=10 \mathrm{~kg}$ Mass of body $B, m_2=20 \mathrm{~kg}$ Total mass of the system, $m=m_1$ $+\mathrm{m}_2=30 \mathrm{~kg}$ Using Newton's second law of motion, the acceleration (a) produced in the system can be calculated as: $\mathrm{F}=\mathrm{ma} \therefore \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{600}{30}=20 \mathrm{~ms}^{-2}$ When force F is applied on body $\mathrm{A}:$
The equation of motion can be written as: $F - T = m_1a$
$\therefore T = F - m_1a = 600 - 10 \times 20 = 400N ......(i)$ When force F is applied on body B:
The equation of motion can be written as: $F - T = m_2a T = F - m_2a$
$\therefore T = 600 - 20 \times 20 = 200N …(ii)$ Which is different from value of T in case (i). Hence our answer depends on which mass end, the force is applied. View full question & answer→Question 185 Marks
A stone of mass $0.25kg$ tied to the end of a string is whirled round in a circle of radius $1.5m$ with a speed of $40 rev./min$ in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of $200N$?
AnswerGiven: Mass of the stone, m = 0.25kg Radius of the circle, r = 1.5m Number of revolution per second, $\text{n}=\frac{40}{60}=\frac{2}{3}\text{rps}$ Angular velocity, $\omega=\frac{\text{v}}{\text{r}}=2\pi\text{n}\ ...(\text{i})$ The centripetal force for the stone is provided by the tension T, in the string, i.e. $T = F_{centripetel} =\frac{\text{mv}^2}{\text{r}}=\text{mr}\omega^2=\text{mr}(2\pi\text{n})^2$
$=0.25\times1.5\times\Big(2\times3.14\times\frac{2}{3}\Big)^2$
$=6.57\text{N}$ Maximum tension in the string, T_{max} = 200N $\text{T}_\text{max}=\frac{\sqrt{\text{mv}^2_\text{max}}}{\text{r}}$ Therefore, $\text{v}=\frac{\sqrt{\text{T}_\text{max}\text{x r}}}{\text{m}}$
$=\frac{\sqrt{200\times1.5}}{0.25}$
$=\sqrt{1200}=34.64\text{m/s}$ Therefore, the maximum speed of the stone is 34.64m/s.
View full question & answer→Question 195 Marks
A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.
AnswerLet, Mass of parent nucleus $=\mathrm{m}$ (At rest) Mass of two daughter nuclei $=\mathrm{m}_1$, and $\mathrm{m}_2$ Initial momentum of the system (parent nucleus) $=0$ Let, Velocity of first daughter nucleus having mass $m_1=v_1$ Velocity of second daughter nucleus having mass $m_2=v_2$ Total linear momentum of the system after disintegration $=\left(m_1 v_1+m_2 v_2\right)$ According to the law of conservation of momentum: Total initial momentum $=$ Total final momentum $0=\left(m_1 \mathrm{v}_1+m_2 \mathrm{v}_2\right) \mathbf{v}_1=\frac{-\mathrm{m}_2 \mathbf{v}_2}{\mathrm{~m}_1}$ Here, the negative sign indicates that the fragments of the parent nucleus move in directions opposite to each other.
View full question & answer→Question 205 Marks
A shell of mass $0.020kg$ is fired by a gun of mass $100kg$. If the muzzle speed of the shell is $80ms^{-1}$, what is the recoil speed of the gun?
AnswerMass of the gun, M= 100kg Mass of the shell, m = 0.020kg Muzzle speed of the shell, v = 80 m/s Recoil speed of the gun = V Both the gun and the shell are at rest initially. Initial momentum of the system = 0 Final momentum of the system = mv - MV Here, the negative sign appears because the directions of the shell and the gun are opposite to each other. According to the law of conservation of momentum: Final momentum = Initial momentum mv - MV = 0 $\therefore\text{V}=\frac{\text{mv}}{\text{M}}$
$=\frac{0.020\times80}{100\times1000}$
$=0.016\text{m/s}$
View full question & answer→Question 215 Marks
A thin circular loop of radius R rotates about its vertical diameter with an angular frequency $\omega.$ Show that a small bead on the wire loop remains at its lowermost point for $\omega\le\sqrt{\frac{\text{g}}{\text{R}}}.$ What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for $\omega=\sqrt{\frac{2\text{g}}{\text{R}}}?$ Neglect friction.
AnswerLet the radius vector joining the bead with the centre make an angle $\theta,$ with the vertical downward direction. 
OP = R = Radius of the circle N = Normal reaction The respective vertical and horizontal equations of forces can be written as: $\text{mg}=\text{N}\cos\theta\ ....(\text{i})$ $\text{ml}\omega=\text{N}\sin\theta\ ...(\text{ii})$ In $\triangle\text{OPQ},$ we have: $\sin\theta=\frac{\text{l}}{\text{R}}$ $\text{l}=\text{R}\sin\theta\ ...(\text{iii})$ Substituting equation (iii) in equation (ii), we get: $\text{m}(\text{R}\sin\theta)\omega^2=\text{N}\sin\theta$ $\text{mR}\omega^2=\text{N}\ ...(\text{iv})$ Substituting equation (iv) in equation (i), we get: $\text{mg}=\text{mR}\omega^2\cos\theta$ $\cos\theta=\frac{\text{g}}{\text{R}\omega^2}\ ...(\text{v})$ Since $\cos\theta\le1,$ the bead will remain at its lowermost point for $\frac{\text{g}}{\text{R}\omega^2\le1},$ i.e., for $\omega\le\Big(\frac{\text{g}}{\text{R}}\Big)^{1/2}$ For $\omega=\Big(\frac{2\text{g}}{\text{R}}\Big)^{1/2}\ \text{or}\ \omega^2=\frac{2\text{g}}{\text{R}}\ ...(\text{vi})$ On equating equations (v) and (vi), we get: $\frac{2\text{g}}{\text{R}}=\frac{\text{g}}{\text{R}\cos\theta}$ $\cos\theta=\frac{1}{2}$ $\therefore\theta=\cos^{-1}(0.5)=60^\circ$ View full question & answer→Question 225 Marks
The driver of a three-wheeler moving with a speed of 36km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0s just in time to save the child. What is the average retarding force on the vehicle ? The mass of the three-wheeler is 400kg and the mass of the driver is 65kg.
AnswerInitial speed of the three-wheeler, u = 36km/h = 10m/s Final speed of the three-wheeler, v = 0m/s Time, t = 4s Mass of the three-wheeler, m = 400kg Mass of the driver, m' = 65kg Total mass of the system, M = 400 + 65 = 465kg Using the first law of motion, the acceleration (a) of the three-wheeler can be calculated as: v = u + at $\therefore\text{a}=\frac{\text{v}-\text{u}}{\text{t}}=\frac{0-10}{4}=-2.5\text{m/s}^2$ The negative sign indicates that the velocity of the three-wheeler is decreasing with time. Using Newton’s second law of motion, the net force acting on the three-wheeler can be calculated as: F = Ma = 465 × (-2.5) = –1162.5N The negative sign indicates that the force is acting against the direction of motion of the three-wheeler.
View full question & answer→Question 235 Marks
A pebble of mass $0.05kg$ is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,
- During its upward motion
- During its downward motion.
- At the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of $45°$ with the horizontal direction? Ignore air resistance.
Answer
- The gravitational force is the only force that acts on the pebble in all three cases. Its magnitude is given by Newton’s second law of motion as:
F = m × a
Where,
F = Net force
m = Mass of the pebble = 0.05kg
$a = g = 10 m/s^2$
$\therefore$ F = 0.05 × 10 = 0.5N
The net force on the pebble in all three cases is $0.5N$ and this force acts in the downward direction.
- If the pebble is thrown at an angle of $45°$ with the horizontal direction, it will have both the horizontal and vertical components of velocity.
- At the highest point, only the vertical component of velocity becomes zero. However, the pebble will have the horizontal component of velocity throughout its motion. This component of velocity produces no effect on the net force acting on the pebble.
View full question & answer→Question 245 Marks
An aircraft executes a horizontal loop at a speed of $720km/h$ with its wings banked at $15°$. What is the radius of the loop?
AnswerSpeed of the aircraft, v = 720km/h $=720\times\frac{5}{18}=200\text{m/s}$ Acceleration due to gravity, $g = 10m/s^2$^ Angle of banking, $\theta=15^\circ$ For radius r, of the loop, we have the relation: $\tan\theta=\frac{\text{v}^2}{\text{rg}}$
$\text{r}=\frac{\text{v}^2}{\text{g}\tan\theta}$
$=\frac{200^2}{(10\times\tan15)}$
$=\frac{4000}{0.26}$
$=14925.37\text{m}$
$=14.92\text{km}$
View full question & answer→Question 255 Marks
Two billiard balls each of mass $0.05kg$ moving in opposite directions with speed $6ms^{-1}$ collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?
AnswerMass of each ball $=0.05 \mathrm{~kg}$ Initial velocity of each ball $=6 \mathrm{~m} / \mathrm{s}$ Magnitude of the initial momentum of each ball, $\mathrm{p}_{\mathrm{i}}=$ $0.3 \mathrm{kgm} / \mathrm{s}$ After collision, the balls change their directions of motion without changing the magnitudes of their velocity. Final momentum of each ball, $\mathrm{p}_{\mathrm{f}}=-0.3 \mathrm{kgm} / \mathrm{s}$ Impulse imparted to each ball = Change in the momentum of the system $=p_f-p_i=-0.3-0.3=-0.6 \mathrm{kgm} / \mathrm{s}$ The negative sign indicates that the impulses imparted to the balls are opposite in direction.
View full question & answer→Question 265 Marks
A body of mass 5kg is acted upon by two perpendicular forces 8N and 6N. Give the magnitude and direction of the acceleration of the body.
AnswerMass of the body, m = 5kg The given situation can be represented as follows: 
The resultant of two forces is given as: $\text{R}=\sqrt{(8)^2+(-6)^2}=\sqrt{64+36}=10\text{N}$ $\theta$ is the angle made by R with the force of 8N $\therefore\theta=\tan^{-1}\Big(\frac{-6}{8}\Big)=-36.87^\circ$ The negative sign indicates that $\theta$ is in the clockwise direction with respect to the force of magnitude 8N. As per Newton’s second law of motion, the acceleration (a) of the body is given as: F = ma $\therefore\text{a}=\frac{\text{F}}{\text{m}}=\frac{10}{5}=2\text{ms}^{-2}$ View full question & answer→Question 275 Marks
You may have seen in a circus a motorcyclist driving in vertical loops inside a ‘deathwell’ (a hollow spherical chamber with holes, so the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required at the uppermost position to perform a vertical loop if the radius of the chamber is 25m?
AnswerThe motorcyclist does not drop down when he is at the uppermost point if centrifugal force on motorcyclist is greater or equal to the weight of motorcyclist. i.e. $\frac{\text{mv}^2}{\text{R}}\ge\text{mg}$ $\Rightarrow\text{v}\ge\sqrt{\text{rg}}$ The minimum speed of motorcyclist not to fall from uppermost point is given by, $\text{v}_\text{min}=\sqrt{\text{rg}}$ Here, given $\text{r}=25,$ $\text{g}=9.8\text{m/s}^2$ $\therefore\text{v}_\text{min}=\sqrt{25\times9.8}=15.7\text{m/s}$
View full question & answer→Question 285 Marks
A train runs along an unbanked circular track of radius 30m at a speed of 54km/h. The mass of the train is $10^6kg$. What provides the centripetal force required for this purpose - The engine or the rails ? What is the angle of banking required to prevent wearing out of the rail?
AnswerRadius of the circular track, r = 30m Speed of the train, v = 54km/h = 15m/s Mass of the train, $m = 10^6kg$ The centripetal force is provided by the lateral thrust of the rail on the wheel. As per Newton’s third law of motion, the wheel exerts an equal and opposite force on the rail. This reaction force is responsible for the wear and rear of the rail. The angle of banking $\theta$, is related to the radius (r) and speed (v) by the relation: $\tan\theta=\frac{\text{v}^2}{\text{rg}}$ $=\frac{15^2}{(30\times10)}$ $=\frac{225}{300}$ $\theta=\tan^{-1}(0.7)=36.87^\circ$ Therefore, the angle of banking is about 36.87°.
View full question & answer→Question 295 Marks
A constant force acting on a body of mass 3.0kg changes its speed from $2.0ms^{-1}$ to $3.5ms^{-1}$ in $25s$. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?
AnswerMass of the body, m = 3kg Initial speed of the body, u = 2m/s Final speed of the body, v = 3.5m/s Time, t = 25s Using the first equation of motion, the acceleration (a) produced in the body can be calculated as: v = u + at $\therefore\text{a}=\frac{(\text{v-u})}{\text{t}}$ $=\frac{(3.5-2)}{25}=0.06\text{ms}^{-2}$ As per Newton’s second law of motion, force is given as: F = ma = 3 × 0.06 = 0.18N Since the application of force does not change the direction of the body, the net force acting on the body is in the direction of its motion.
View full question & answer→Question 305 Marks
Figure. shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with $1ms^{-2}$. What is the net force on the man? If the coefficient of static friction between the man’s shoes and the belt is $0.2$, up to what acceleration of the belt can the man continue to be stationary relative to the belt? (Mass of the man = $65kg$.)
AnswerMass of the man, m = 65kg Acceleration of the belt, $a = 1m/s^2$ Coefficient of static friction, μ = 0.2 The net force F, acting on the man is given by Newton’s second law of motion as: $F_{net} = ma = 65 × 1 = 65N$ The man will continue to be stationary with respect to the conveyor belt until the net force on the man is less than or equal to the frictional force fs, exerted by the belt, i.e., $F’_{net} = f_s ma’ = \mu mg $
$\therefore a‘ = 0.2 \times 10 = 2m/s^2$ Therefore, the maximum acceleration of the belt up to which the man can stand stationary is $2m/s^2$.
View full question & answer→Question 315 Marks
A $70kg$ man stands in contact against the inner wall of a hollow cylindrical drum of radius $3m$ rotating about its vertical axis with $200rev/min$. The coefficient of friction between the wall and his clothing is $0.15$. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?
AnswerMass of the man, m = 70kg Radius of the drum, r = 3m Coefficient of friction, $\mu=0.15$ Frequency of rotation, $\text{v}=200\text{rev/min}=\frac{200}{60}=\frac{10}{3}\text{rev/s}$ The necessary centripetal force required for the rotation of the man is provided by the normal force $(F_N)$. When the floor revolves, the man sticks to the wall of the drum. Hence, the weight of the man (mg) acting downward is balanced by the frictional force $(\text{f}=\mu\text{F}_\text{N})$ acting upward. Hence, the man will not fall until: $\text{mg}<\text{f}$ $\text{mg}<\mu\text{F}_\text{N}=\mu\text{mr}\omega^2$ $\text{g}<\mu\text{r}\omega^2$ $\omega>\Big(\frac{\text{g}}{\mu\text{r}}\Big)^{1/2}$ The minimum angular speed is given as: $\omega_\text{min}>\Big(\frac{\text{g}}{\mu\text{r}}\Big)^{1/2}$ $=\Big(\frac{10}{0.15\times3}\Big)^{1/2}=4.71\text{ rad s}^{-1}$
View full question & answer→Question 325 Marks
An aircraft executes a horizontal loop at a speed of 720km/h with its wings banked at 15°. What is the radius of the loop?
AnswerSpeed of the aircraft, v = 720km/h $=720\times\frac{5}{18}=200\text{m/s}$ Acceleration due to gravity, $g = 10m/s^2$ Angle of banking, $\theta=15^\circ$ For radius r, of the loop, we have the relation: $\tan\theta=\frac{\text{v}^2}{\text{rg}}$ $\text{r}=\frac{\text{v}^2}{\text{g}\tan\theta}$ $=\frac{200^2}{(10\times\tan15)}$ $=\frac{4000}{0.26}$ $=14925.37\text{m}$ $=14.92\text{km}$
View full question & answer→Question 335 Marks
Two billiard balls each of mass $0.05kg$ moving in opposite directions with speed $6ms^{-1}$ collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?
AnswerMass of each ball $=0.05 \mathrm{~kg}$ Initial velocity of each ball $=6 \mathrm{~m} / \mathrm{s}$ Magnitude of the initial momentum of each ball, $\mathrm{p}_{\mathrm{i}}=$ $0.3 \mathrm{kgm} / \mathrm{s}$ After collision, the balls change their directions of motion without changing the magnitudes of their velocity. Final momentum of each ball, $\mathrm{p}_{\mathrm{f}}=-0.3 \mathrm{kgm} / \mathrm{s}$ Impulse imparted to each ball = Change in the momentum of the system $=p_f-p_i=-0.3-0.3=-0.6 \mathrm{kgm} / \mathrm{s}$ The negative sign indicates that the impulses imparted to the balls are opposite in direction.
View full question & answer→Question 345 Marks
Two masses of 5 kg and 3 kg are suspended with help of massless inextensible strings as shown in Calculate $\mathrm{T}_1$ and $\mathrm{T}_2$ when whole system is going upwards with acceleration $=2 \mathrm{~m} \mathrm{~s}^2\left(\mathrm{use} \mathrm{~g}=9.8 \mathrm{~m} \mathrm{~s}^{-2}\right)$.
AnswerAs the whole system is going up with acceleration = a = $2ms^{-2} \text{m}_1=5\text{kg}\ \text{m}_2=3\text{kg}\ \text{g}=9.8\text{m/ s}^2$
Tension in a string is equal and opposite in all parts of a string.
Forces on mass $m_1$
$\text{T}_1-\text{T}_2-\text{m}_1\text{g}=\text{m}_1\text{a}$
$\text{T}_1-\text{T}_2-5\text{g}+5\text{a}$
$\text{T}_1-\text{T}_2=5\text{g}+5\text{a}$
$\text{T}_1-\text{T}_2=5(9.8+2)$
$=5\times11.8$
$\text{T}_1-\text{T}_2=59.0\ \text{N}$
Forces on mass $m_2$
$\text{T}_2-\text{m}_2\text{g}=\text{m}_2\text{a}$
$\text{T}_2=\text{m}(\text{g}+\text{a})=3(9.8+2)=3\times11.8$
$\text{T}_2=35.4$
$\text{T}_1=\text{T}_2+59.0\Rightarrow\text{T}_1=53.4+59.0=94.4\ \text{N}$ View full question & answer→Question 355 Marks
A body of mass 5kg is acted upon by two perpendicular forces 8N and 6N. Give the magnitude and direction of the acceleration of the body.
AnswerMass of the body, m = 5kg The given situation can be represented as follows: 
The resultant of two forces is given as: $\text{R}=\sqrt{(8)^2+(-6)^2}=\sqrt{64+36}=10\text{N}$ $\theta$ is the angle made by R with the force of 8N $\therefore\theta=\tan^{-1}\Big(\frac{-6}{8}\Big)=-36.87^\circ$ The negative sign indicates that $\theta$ is in the clockwise direction with respect to the force of magnitude 8N. As per Newton’s second law of motion, the acceleration (a) of the body is given as: F = ma $\therefore\text{a}=\frac{\text{F}}{\text{m}}=\frac{10}{5}=2\text{ms}^{-2}$ View full question & answer→Question 365 Marks
An object weighing $70kg$ is kept in a lift. Find its weight as recorded by a spring balance when the lift
- Moves upwards with a uniform velocity of $5ms^{-1}$,
- Moves upwards with a uniform acceleration of $2.2ms^{-2}$
- Moves downwards with a uniform acceleration of $2.8ms^{-2}$ and
- Falls freely under gravity.
Answer
- When the lift is moving upwards with a uniform velocity $5ms^{-1}$ acceleration is zero), the reaction R or the pressure on the base is,
R = mg = 70 x 9.8N = 686N
- When the lift is moving upwards with a uniform acceleration of $2.2ms^{-2}$, the reaction R' or the pressure on the base increases and is given by,
R' = m (g + a) = 70(9.8 + 2.2)N = 840N
- When the lift descends with a uniform acceleration of $2.8ms^{-2}$, the reaction R" is given by,
R" = m(g - a) = 70(9.8 - 2.8)N = 490N
- When the lift falls freely under gravity, the reaction R" is given by,
R'' = m(g - g) = 0
i.e., the object appears to have become weightless. View full question & answer→Question 375 Marks
The velocity of a body of mass 2kg as a function of t is given by $\text{v}(\text{t})=2\text{t}\hat{\text{i}}+\text{t}^2\hat{\text{j.}}$ Find the momentum and the force acting on it, at time $\text{t}=2\text{s}.$
Answer$\text{m}=2\text{kg}$ $\vec{\text{v}}\text{(t)}=2\text{t}\hat{\text{i}}+\text{t}^2\hat{\text{j.}}$ $\vec{\text{v}}\ \text{at}2\sec,\vec{\text{v}}(2)=2(2)\hat{\text{i}}+(2)^2\hat{\text{j}},\text{v}(2)=4\hat{\text{i}}+4\hat{\text{j}}$ Momentum $\vec{\text{p}}(2)=\text{m}\vec{\text{v}}(2)$ $\vec{\text{p}}(2)=2\Big[4 \hat{\text{i}}+4\hat{\text{j}}\Big],\text{p}(2)=8\hat{\text{i}}+8\hat{\text{j}}\text{kg}\ \text{ms}^{-1}$ $\vec{\text{F}}=\text{m}\vec{\text{a}}$ $\vec{\text{F}}(2)=\text{m}\vec{\text{a}}(2)$ $\vec{\text{v}}\text{(t)}=2\text{t}\hat{\text{i}}+\text{t}^2\hat{\text{j}}$ $\hat{\text{a}}\text{(t)}=\frac{\text{d}\vec{\text{v}\text{(t)}}}{\text{dt}}=2\hat{\text{i}}+2\text{t}\hat{\text{j}}$ $\hat{\text{a}}(2)=2\hat{\text{i}}+2(2)\hat{\text{j}}=2\hat{\text{i}}+4\hat{\text{j}}$ $\therefore\vec{\text{F}}(2)=2(2\hat{\text{i}}+4\hat{\text{j}}=4\hat{\text{i}}+8\hat{\text{j}}\text{N}$
View full question & answer→Question 385 Marks
You may have seen in a circus a motorcyclist driving in vertical loops inside a ‘deathwell’ (a hollow spherical chamber with holes, so the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required at the uppermost position to perform a vertical loop if the radius of the chamber is 25m?
AnswerThe motorcyclist does not drop down when he is at the uppermost point if centrifugal force on motorcyclist is greater or equal to the weight of motorcyclist. i.e. $\frac{\text{mv}^2}{\text{R}}\ge\text{mg}$ $\Rightarrow\text{v}\ge\sqrt{\text{rg}}$ The minimum speed of motorcyclist not to fall from uppermost point is given by, $\text{v}_\text{min}=\sqrt{\text{rg}}$ Here, given $\text{r}=25,$ $\text{g}=9.8\text{m/s}^2$ $\therefore\text{v}_\text{min}=\sqrt{25\times9.8}=15.7\text{m/s}$
View full question & answer→Question 395 Marks
A train runs along an unbanked circular track of radius $30m$ at a speed of $54km/h$. The mass of the train is $10^6kg$. What provides the centripetal force required for this purpose - The engine or the rails ? What is the angle of banking required to prevent wearing out of the rail?
AnswerRadius of the circular track, $r = 30m$ Speed of the train, $v = 54km/h = 15m/s$ Mass of the train, $m = 10^6kg$ The centripetal force is provided by the lateral thrust of the rail on the wheel. As per Newton’s third law of motion, the wheel exerts an equal and opposite force on the rail. This reaction force is responsible for the wear and rear of the rail. The angle of banking $\theta$, is related to the radius (r) and speed (v) by the relation: $\tan\theta=\frac{\text{v}^2}{\text{rg}}$ $=\frac{15^2}{(30\times10)}$ $=\frac{225}{300}$ $\theta=\tan^{-1}(0.7)=36.87^\circ$ Therefore, the angle of banking is about 36.87°.
View full question & answer→Question 405 Marks
A body $m_1$ of mass 9kg and another body $m_2$ of mass 6kg are connected by a light inextensible string. Consider a smooth inclined plane of inclination $30^\circ$ over which one of them can be placed while the other hangs vertically and freely. Show that $m_1$ will drag $m_2$ up the whole length of the plane in half the time that $m_2$ hanging vertically would take to draw $m_1$ up the plane.
Answer
Case I: Let $a_1$_ be the acceleration of the system when 9kg mass hangs freely and T the tension in the string. $\text{M}_1\text{g}-\text{T}=\text{m}_1\text{a}_1$
$\text{T}-\text{m}_2\text{g}\sin30^\circ=\text{m}_2\text{a}_1$
$\Rightarrow\text{g}(\text{m}_1-\text{m}_2\sin30^\circ)$
$=\text{a}_1(\text{m}_1+\text{m}_2)$
$\Rightarrow\text{a}_1=\frac{\text{g}\Big(9-6\times\frac{1}{2}\Big)}{15}$
$=\frac{\text{6g}}{15}=\frac{\text{2g}}{5}$ Case II: Let a_2 be the acceleration of the system when 6kg mass hangs freely and T’ the tension in the string. $\text{m}_2\text{g}-\text{T}=\text{m}_\text{2}\text{a}_2$
$\text{T}-\text{m}_1\text{g}\sin30^\circ=\text{m}_1\text{a}_2$
$\Rightarrow\text{g}(\text{m}_2-\text{m}_1\sin30^\circ)=\text{a}_2(\text{m}_2+\text{m}_1)$
$\Rightarrow\text{g}\Big(6-9\times\frac{1}{2}\Big)$
$=\text{a}_2(6+9)\Rightarrow\text{a}_2=\frac{3\text{g}}{30}=\frac{\text{g}}{10}$ If s is the lenght of the plane, In case (i), $\text{S}=\frac{1}{2}\text{a}_1\text{t}_1^2$ In case (ii), $\text{S}=\frac{1}{2}\text{a}_1\text{t}^2_2$
$\Rightarrow\text{a}_1\text{t}^2_1=\text{a}_2\text{t}_2^2$
$\Rightarrow\frac{\text{t}_1}{\text{t}_2}=\sqrt{\frac{\text{a}_2}{\text{a}_1}}=\sqrt{\frac{\frac{\text{g}}{10}}{\frac{2\text{g}}{5}}}$
$=\sqrt{\frac{1}{4}}$
$\Rightarrow\text{t}_1:\text{t}_2=1:2.$ View full question & answer→Question 415 Marks
shows (x, t), (y, t) diagram of a particle moving in 2-dimensions.
If the particle has a mass of 500g, find the force (direction and magnitude) acting on the particle.
AnswerFrom graph (a) $\text{v}_\text{x}=\frac{\text{dx}}{\text{dt}}=\frac{2}{2}=1\text{ms}^{-1}$ $\text{a}_\text{x}=\frac{\text{d}^2\text{x}}{\text{dt}^2}=\frac{\text{dv}_\text{x}}{\text{dt}}=0$ From figure (b) $\text{y}=\text{t}^2$ $\text{v}_\text{y}=\frac{\text{dy}}{\text{dt}}=2\text{t}$ $\text{a}_\text{y}=\frac{\text{dv}_\text{y}}{\text{dt}}=2$$\therefore\text{F}_\text{y}=\text{ma}_\text{y}\ \text{m}=500\text{g}=.5\text{kg}$
$\text{F}_\text{y}=.5\times2=1\text{N}$ toward $\text{Y}-\text{axis}$$\text{F}_\text{x}=.5\times0=0\text{N}$
$\text{F}=\sqrt{\text{F}_\text{x}^2+\text{F}_\text{y}^2}=\sqrt{0^2+1^2}$
F = N toward Y-axis.
View full question & answer→Question 425 Marks
There are four forces acting at a point P produced by strings as shown in which is at rest. Find the forces $F_1$ and $F_2$. 
AnswerAs the particle is rest or a = 0. So resultant force due to all forces will be zero. $\therefore$ Net components along X and Y-axis will be zero. Resolving all forces along X-axis $\text{F}_\text{x}=0$ $\text{F}_1+1\cos45^\circ-2\cos45^\circ=0$ or $\text{F}_1-1\cos45^\circ=0$ $\text{F}_1=\cos45^\circ=\frac{1}{\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}=\frac{1.414}{2}=0.707\text{N}$ Resolving all forces along Y-axis $\text{F}_\text{y}=0$ $-\text{F}_2+1\cos45^\circ+2\cos45%\circ=0$ $-\text{F}_2=-3\cos45^\circ$ $\text{F}_2=3.\frac{1}{\sqrt{2}}=\frac{3\sqrt{2}}{2}=\frac{3\times1.414}{2}=3\times0.707=2.121\text{N.}$
View full question & answer→Question 435 Marks
There are three forces $F_1, F_2$ and $F_3$ acting on a body, all acting on a point P on the body. The body is found to move with uniform speed.
- Show that the forces are coplanar.
- Show that the torque acting on the body about any point due to these three forces is zero.
Answer
- As the body is moving with uniform speed after the action of three forces $\vec{\text{F}}_1\vec{\text{F}}_2$ and $\vec{\text{F}}_3$ on a point on body, so the acceleration of body is zero (as no circular motion)
$\therefore$ F = ma, so the resultant force due to $\vec{\text{F}}_1+\vec{\text{F}}_2+\vec{\text{F}}_3=0$
$\vec{\text{F}}_1+\vec{\text{F}}_2=-\vec{\text{F}}_3$ or $\vec{\text{F}}_3=-(\vec{\text{F}}_1+\vec{\text{F}}_2)$
Consider the forces $\vec{\text{F}}_1$ or $\vec{\text{F}}_2$ are in the plane of the paper, the resultant of $\vec{\text{F}}_1$ and $\vec{\text{F}}_2$ will also be on the same plane of the paper, in $\Big[-(\vec{\text{F}}_1+\vec{\text{F}}_2)\Big]$ only direction is reverse on the same plane. As $\vec{\text{F}}_3=-(\vec{\text{F}}_1+\vec{\text{F}}_2).$ so $\vec{\text{F}}_3$ will be in the same plane i.e., $\vec{\text{F}}_1, \vec{\text{F}}_2$ and $\vec{\text{F}}_3$ are coplanar.
- As the resultant of $\vec{\text{F}}_1,\ \vec{\text{F}}_2$ and $\vec{\text{F}}_3$ is zero and Torque $\text{r}\times\vec{\text{F}}=0$
as $\vec{\text{F}}=\vec{\text{F}}_1+\vec{\text{F}}_2+\vec{\text{F}}_3=0$
So torque acting on a body at any point will be always zero. View full question & answer→Question 445 Marks
A constant force acting on a body of mass $3.0kg$ changes its speed from $2.0ms^{-1}$ to $3.5ms^{-1}$ in $25s$. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?
AnswerMass of the body, m = 3kg Initial speed of the body, u = 2m/s Final speed of the body, v = 3.5m/s Time, t = 25s Using the first equation of motion, the acceleration (a) produced in the body can be calculated as: v = u + at $\therefore\text{a}=\frac{(\text{v-u})}{\text{t}}$ $=\frac{(3.5-2)}{25}=0.06\text{ms}^{-2}$ As per Newton’s second law of motion, force is given as: F = ma = 3 × 0.06 = 0.18N Since the application of force does not change the direction of the body, the net force acting on the body is in the direction of its motion.
View full question & answer→Question 455 Marks
The displacement vector of a particle of mass m is given by $\text{r}\text{(t})=\hat{\text{i}}\text{A}\cos\omega\text{t}+\hat{\text{j}}\text{B}\sin\omega\text{t}.$Show that $\text{F}=-\text{m}\omega^2\text{r}.$
AnswerThe Main concept used: To plot the graph (r - t) or trajectory we relate x and y coordinates. $\text{x}=\text{A}\cos\omega\text{t}$ (Given) $\Rightarrow\text{v}_\text{x}=\frac{\text{dx}}{\text{dt}}=-\text{A}\omega\sin\omega\text{t}$ $\Rightarrow\text{a}_\text{x}=\frac{\text{dv}_\text{z}}{\text{dt}}=-\text{A}\omega^2\cos\omega\text{t}$ And $\text{y}\text{B}\sin\omega\text{t}$ (Given) $\Rightarrow\text{v}_\text{y}=\frac{\text{dy}}{\text{dt}}=\text{B}\omega\cos\omega\text{t}$ $\text{a}_\text{y}=\frac{\text{dy}_\text{y}}{\text{dt}}=-\text{B}\omega^2\sin\omega\text{t}$ $\text{a}=\text{a}_\text{x}\hat{\text{i}}+\text{a}_\text{y}\hat{\text{j}}$ $=-\hat{\text{i}}\text{A}\text{w}^2\cos\text{w}\text{t}-\hat{\text{j}}\text{B}\omega^2\sin\omega\text{t}$ $=-\omega^2[\hat{\text{i}}\text{A}\cos\omega\text{t}+\hat{\text{j}}\text{B}\sin\omega\text{t}]$ $\text{a}=\omega^2\text{r}(\text{t})$ $\therefore$ Force acting on particle =ma = - m $\omega^2\vec{\text{r}}-(\text{t})$
View full question & answer→Question 465 Marks
Figure. shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with $1ms^{-2}$. What is the net force on the man? If the coefficient of static friction between the man’s shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt? (Mass of the man = 65kg.)
AnswerMass of the man, $\mathrm{m}=65 \mathrm{~kg}$ Acceleration of the belt, $a=1 \mathrm{~m} / \mathrm{s}^2$ Coefficient of static friction, $\mu=0.2$ The net force $F$, acting on the man is given by Newton's second law of motion as: $F_{\text {net }}=m a=65 \times 1=65 \mathrm{~N}$ The man will continue to be stationary with respect to the conveyor belt until the net force on the man is less than or equal to the frictional force fs, exerted by the belt, i.e., $\mathrm{F}_{\text {net }}=\mathrm{f}_{\mathrm{s}} \mathrm{ma}^{\prime}=\mu \mathrm{mg} \therefore \mathrm{a}^{\prime}=0.2 \times 10=2 \mathrm{~m} / \mathrm{s}^2$ Therefore, the maximum acceleration of the belt up to which the man can stand stationary is $2 \mathrm{~m} / \mathrm{s}^2$.
View full question & answer→Question 475 Marks
A $70kg$ man stands in contact against the inner wall of a hollow cylindrical drum of radius $3m$ rotating about its vertical axis with $200rev/min$. The coefficient of friction between the wall and his clothing is $0.15$. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?
AnswerMass of the man, m = 70kg Radius of the drum, r = 3m Coefficient of friction, $\mu=0.15$ Frequency of rotation, $\text{v}=200\text{rev/min}=\frac{200}{60}=\frac{10}{3}\text{rev/s}$ The necessary centripetal force required for the rotation of the man is provided by the normal force $(F_N)$.
When the floor revolves, the man sticks to the wall of the drum.
Hence, the weight of the man (mg) acting downward is balanced by the frictional force $(\text{f}=\mu\text{F}_\text{N})$ acting upward.
Hence, the man will not fall until: $\text{mg}<\text{f}$ $\text{mg}<\mu\text{F}_\text{N}=\mu\text{mr}\omega^2$ $\text{g}<\mu\text{r}\omega^2$ $\omega>\Big(\frac{\text{g}}{\mu\text{r}}\Big)^{1/2}$ The minimum angular speed is given as: $\omega_\text{min}>\Big(\frac{\text{g}}{\mu\text{r}}\Big)^{1/2}$ $=\Big(\frac{10}{0.15\times3}\Big)^{1/2}=4.71\text{ rad s}^{-1}$
View full question & answer→Question 485 Marks
A girl riding a bicycle along a straight road with a speed of $5m s^{–1}$ throws a stone of mass $0.5kg$ which has a speed of $15m s^{–1}$ with respect to the ground along her direction of motion. The mass of the girl and bicycle is $50kg$. Does the speed of the bicycle change after the stone is thrown? What is the change in speed, if so?
AnswerGirl and cycle Body $\text{m}_1=50\text{kg}\ \ \ \text{m}_2=0.5\text{kg}$
$\mu_1=5\text{m/ s}$ forward $\mu_2=5\ \text{m/ s}$ forward $\text{v}_1=?\ \ \text{v}_2=15\text{m/ s}$ forward According to the law of conservation of momentum. Initial momentum (Girl, cycle, body) = Final momentum (cycle + Girl) and body $(\text{m}_1+\text{m}_2)\mu_1=\text{m}_1\text{v}_1+\text{m}_2\text{v}_2$
$(50+0.5)\times5+50\times\text{v}_1+0.5\times15$
$50.5\times5-7.5=50\text{v}_1$
$50\text{v}_1=252.5-7.5=245.0$
$\text{v}_1\frac{245.0}{50}=4.9\text{m/ s}$ Hence, the speed of cycle and girl decreased by 5 - 4.9 = 0.1m/ s
View full question & answer→Question 495 Marks
State Newton's third law of motion. Discuss its consequences.
AnswerNewton's third law of motion states that for any action, there is equal and opposite reaction. So, if a body applies a force $F_{12}$ on body 2 (action), then body 2 also applies a force $F_{21}$ on body 1 but in opposite direction, then $F_{21} = - F_{12}$ In terms of magnitude $|F_{21} | = | F_{12}|$ It is very important to note that $F_{12}$ and $F_{21}$, though are equal in magnitude and opposite in direction yet act on different points or else no motion will be possible. For example, hands pull up a chest expander (spring) and spring in turn exerts force on the arms. A football pressed reacts on the foot with the same force and so on. The most important consequence of the third law of motion is the law of motion is the law of conservation of linear momentum and its application in collision problems. Since $\text{F}_{12}=\text{F}_{21}$ and $\text{F}=\text{m}\frac{\Delta\upsilon}{\Delta\text{t}}$ $\therefore\text{m}_1\frac{\Delta\upsilon_1}{\Delta\text{t}}=-\text{m}_2\frac{\Delta\upsilon_2}{\Delta\text{t}}$ Here $\Delta\text{t}$ is the time for which the bodies come in contact during impact. This is same for the two bodies of masses $m_1$ and $m_2$ and having velocity changes $\Delta\text{v}_1$ and $\Delta\text{v}_2$ respectively. Let $\text{u}_1,\text{u}_2$ and $\text{v}_1,\text{v}_2$ be the initial and final velocities of the two masses before and after collision, then, $\text{m}_1(\text{v}_1-\text{u}_1)=\text{m}_2(\text{v}_2-\text{u}_2)$ or $\text{m}_1\text{u}_1+\text{m}_2\text{u}_2=\text{m}_1\text{v}_1+\text{m}_2\text{v}_2$ Momentum before impact = Momentum after impact. (This is known as the law of conservation of momentum).
View full question & answer→Question 505 Marks
Assuming the length of a chain to be L and coefficient of static friction $\mu,$ calculate the maximum length of the chain which can be held outside a table without sliding.
Answer
Let y be the maximum length of the chain, which can be held outside the table without sliding. Lenth of the chain on the table = (L - y) Weight of this part of chain, $\text{W}=\frac{\text{M}}{\text{L}}(\text{L}-\text{y})\text{g}$ Weight of hanging part of chain $=\text{W}=\frac{\text{M}}{\text{L}}\text{y g}$ For equilibrium according to fig. force of friction (f) = wt. of hanging part of chain $\mu\text{R}=\text{W}$ $\mu\text{W}=\text{W}$ $\mu.\frac{\text{M}}{\text{L}}(\text{L}-\text{y})\text{gs}=\frac{\text{M}}{\text{L}}\text{y g}$ $\mu\text{mg}-\mu\frac{\text{M y g}}{\text{L}}=\frac{\text{M}}{\text{L}}\text{yg}$ $\mu\text{Mg}=\frac{\text{M}}{\text{L}}\text{y g}(\text{l}+\mu)$ $\text{y}=\frac{\mu\text{L}}{(\text{l}+\mu)}$ View full question & answer→