Question 13 Marks
A man of mass $70kg$ stands on a weighing scale in a lift which is moving. Upwards with a uniform acceleration of $5ms^{-2}$. What would be the readings on the scale in each case?
AnswerMass of the man, m = 70kg Acceleration, $a = 5m/s^2$ upward Using Newton’s second law of motion, we can write the equation of motion as: R - mg = ma R = m(g + a) = 7(10 + 5) = 70 × 15 = 1050N
$\therefore$ Reading on the weighing scale $=\frac{1050}{\text{g}}=\frac{1050}{10}=105\text{kg}$
View full question & answer→Question 23 Marks
A constant retarding force of $50N$ is applied to a body of mass $20kg$ moving initially with a speed of $15ms^{-1}$. How long does the body take to stop?
AnswerRetarding force, F = –50N Mass of the body, m = 20kg Initial velocity of the body, u = 15m/s Final velocity of the body, v = 0 Using Newton’s second law of motion, the acceleration (a) produced in the body can be calculated as: F = ma –50 = 20 × a
$\therefore\text{a}=\frac{-50}{20}=-2.5\text{ms}^{-2}$ Using the first equation of motion, the time (t) taken by the body to come to rest can be calculated as: v = u + at $\therefore\text{t}=\frac{-\text{u}}{\text{a}}=\frac{-15}{-2.5}=6\text{s}$
View full question & answer→Question 33 Marks
A man of mass $70kg$ stands on a weighing scale in a lift which is moving. Downwards with a uniform acceleration of $5ms^{-2}$.
AnswerMass of the man, $m = 70kg$ Acceleration, $a = 5m/s^2$ downward Using Newton’s second law of motion, we can write the equation of motion as: R + mg = ma R = m(g - a) = 70(10 - 5) = 70 × 5 = 350N
$\therefore$ Reading on the weighing scale $=350\text{g}=\frac{350}{10}=35\text{kg}$
View full question & answer→Question 43 Marks
A bob of mass $0.1\ kg$ hung from the ceiling of a room by a string $2m$ long is set into oscillation. The speed of the bob at its mean position is $1ms^{-1}$ . What is the trajectory of the bob if the string is cut when the bob is $(a)$ at one of its extreme positions, $(b)$ at its mean position.
Answer
- Vertically downward: At the extreme position, the velocity of the bob becomes zero. If the string is cut at this moment, then the bob will fall vertically on the ground.
- Parabolic path: At the mean position, the velocity of the bob is $1m/s$. The direction of this velocity is tangential to the arc formed by the oscillating bob. If the bob is cut at the mean position, then it will trace a projectile path having the horizontal component of velocity only. Hence, it will follow a parabolic path.
View full question & answer→Question 53 Marks
A man of mass $70kg$ stands on a weighing scale in a lift which is moving. Upwards with a uniform speed of $10 ms^{-1}$.
AnswerMass of the man, m = 70kg Acceleration, a = 0 Using Newton’s second law of motion, we can write the equation of motion as: R - mg = ma Where, ma is the net force acting on the man. As the lift is moving at a uniform speed, acceleration a = 0 $\therefore$ R = mg = 70 × 10 = 700N
$\therefore$ Reading on the weighing scale $=\frac{700}{\text{g}}=\frac{700}{10}=70\text{kg}$
View full question & answer→Question 63 Marks
A rocket with a lift-off mass $20,000kg$ is blasted upwards with an initial acceleration of $5.0ms^{-2}$. Calculate the initial thrust (force) of the blast.
AnswerGiven: Mass of the rocket, $\mathrm{m}=20,000 \mathrm{~kg}$ Initial acceleration, $a=5 \mathrm{~m} / \mathrm{s}^2$ Acceleration due to gravity, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ Using Newton's second law of motion, the net force (thrust) acting on the rocket is given by the relation: ( $F-\mathrm{mg}$ ) = ma $\mathrm{F}=$ $m(g+a)=(20000 \times(10+5))=(20000 \times 15)=3 \times 10^5 \mathrm{~N}$
View full question & answer→Question 73 Marks
A man of mass 70kg stands on a weighing scale in a lift which is moving. What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?
AnswerWhen the lift moves freely under gravity, acceleration a = g Using Newton’s second law of motion, we can write the equation of motion as: R + mg = ma R = m(g – a) = m(g - g) = 0 $\therefore$ Reading on the weighing scale $\frac{0}{\text{g}}=0\text{kg}$ The man will be in a state of weightlessness.
View full question & answer→MCQ 83 Marks
One end of a string of length $l$ is connected to a particle of mass $m$ and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed $v$ the net force on the particle $($directed towards the centre$)$ is : $T$ is the tension in the string. $[$Choose the correct alternative$].$
AnswerCorrect option: A. $\text{T}.$
The net force $T$ on the particle is directed towards the centre. When a particle connected to a string revolves in a circular path around a centre,
the centripetal force is provided by the tension produced in the string. Hence, in the given case, the net force on the particle is the tension $T$, i.e.
$\text{F}=\text{T}=\frac{\text{m}\nu^2}{\text{l}}$; Where $F$ is the net force acting on the particle
View full question & answer→Question 93 Marks
Two blocks $3kg$ and $2kg$ are suspended from a rigid support by two inextensible wires, each of length $1m$ and having linear mass density $0.2kg/ m$. Find the tension at the mid-point of each wire as the arrangement gets an upward acceleration of $2m/ s^2$.
AnswerSince the strings carry mass there will be varying tension at different points. At the middle of the string attached to the roof, the forces are as shown.
The tension $T_A$ when the array is at rest is, $\text{T}_\text{A}=\frac{1}{2}(0.2)+5+(0.2)$
$\Rightarrow\ \text{T}_\text{A}=5.3\text{kg}$
The tension $T_B$ when the array stays at rest is,
$T_B = 2kg + 0.1kg = 2.1kg$
At the middle of the lower string the forces are
When the arrangement is accelerated, $T_A = 5.3(g + a) = 62.54N$ and $T_B = (2.1)(g + a) = 24.78N$ View full question & answer→Question 103 Marks
A table with smooth horizontal surface is fixed in a cabin that rotates with angular speed o in a circular path of radius R. A smooth groove AB of length L(< < R) is made on the surface of table as shown in figure.
A small particle is kept at the point A in the groove and is released to move, find the time taken by the particle to reach the point B.
AnswerLet us analyse the motion of particle with respect to table which is moving with cabin with an angular speed of $\omega.$ Along AB centrifugal force of magnitude $\text{m}\omega^2\text{R}$ will act at A on the particle which can be treated as constant from A to B as L << R. $\therefore$ Acceleration of particle along AB with respect to cabin $\text{a}=\omega^2\text{R}$ (constant) Required time 't’ is given by, $\text{S}=\text{ut}+\frac{1}{2}\text{at}^2$ $\Rightarrow\text{L}=0+\frac{1}{2}\times\omega^2\text{Rt}^2$ $\Rightarrow\text{t}=\sqrt{\frac{2\text{L}}{\omega^2\text{R}}}$
View full question & answer→Question 113 Marks
A cricket ball of mass $150g$ is moving with a velocity of $12ms^{-1}$, and is hit by a bat, so that the ball is turned back with a velocity of $20ms^{-1}$. The force of the blow acts for $0.01$ second on the ball. Find the average force exerted by the bat on the ball.
AnswerChange in momentum $=150 \times 10^{-3}[20-12]=1200 \times 10^{-3} \mathrm{~kg} \mathrm{~ms}^{-1}$ Since $t=0.01$ second
$\therefore\ \text{Average force}=\frac{1200\times10^{-3}}{10^{-2}}=120\text{N}$
View full question & answer→Question 123 Marks
A man of mass $70kg$ stands on a weighing scale in a lift which is moving. Upwards with a uniform acceleration of $5ms^{-2}$. What would be the readings on the scale in each case?
AnswerMass of the man, $m = 70kg$ Acceleration, $a = 5m/s^2$ upward Using Newton’s second law of motion, we can write the equation of motion as: R - mg = ma R = m(g + a) = 7(10 + 5) = 70 × 15 = 1050N
$\therefore$ Reading on the weighing scale $=\frac{1050}{\text{g}}=\frac{1050}{10}=105\text{kg}$
View full question & answer→Question 133 Marks
A helicopter of mass $500\ kg$ rises with a vertical acceleration of $10ms ^{-2}$. The weight of pilot is $60\ kg$. Give the magnitude and direction of :
- force on the floor of the helicopter by the pilot
- action of the rotor of the helicopter on the surrounding air
- force on the helicopter due to the surrounding air.
$($Take $g = 10 ms^{-2}).$ Answer$i.$ Force on the floor by the pilot
$=m g+m a=m(8+a)=60(10+10)=1200 N($downward$)$
$ii.$ Force of helicopter on the surrounding air $=\left(m_1+m_2\right)(g+a)$
$=(500+60)(10+10)=11200 N ($downwards$)$
$iii.$ According to Newton's third law of motion, action and reaction are equal and opposite.
$\therefore$ Force on the helicopter due to surrounding air $= 11200N ($upwards$)$.
View full question & answer→Question 143 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$\mathrm{u}=0, \mathrm{a}=2 \mathrm{~ms}^{-2}, \mathrm{t}=10 \mathrm{sec} ., \mathrm{h}=6 \mathrm{~m}$ At $\mathrm{t}=10 \mathrm{sec}$, the velocity of the truck $=\mathrm{v}_{10}=0+2 \times 10=20 \mathrm{~ms}^{-1}$
$a.$ At $t=11 \mathrm{sec}$, the stone will have horizontal velocity of $u_x=20 \mathrm{~ms}^{-1}$
and vertical velocity
$\mathrm{u}_{\mathrm{y}}=0+\mathrm{g} \times 1$
$=10 \mathrm{~ms}^{-1}$
$\therefore$ Velocity of stone at $t=11 \mathrm{~s}$ will be
$\sqrt{\text{u}^2_\text{x}+\text{u}^2_\text{y}}=\sqrt{20^2+10^2}$
$=\sqrt{500}=10\sqrt{5}\text{ ms}^{-1}$
$b.$ Acceleration of stone at $t = 11 \text{sec}$ will be equal to the acceleration due to gravity.
View full question & answer→Question 153 Marks
Name a varying mass system. Derive an expression for velocity of propulsion of a rocket at any instant.
AnswerRocket propelled up in space.Let u be the velocity of exhaust gas (burnt) and $\frac{\text{dM}}{\text{dt}}$ be the rate at which fuel is burnt.
Let M be the mass at any instant and dv be the change in velocity of the rocket at that instant.
From conservation of momentum, we have,
Change in momentum of rocket = Change in momentum of burnt gases.
i.e., Mdv = -udM
$\therefore\ \text{dv}=-\frac{\text{udM}}{\text{M}}$Integrating we get, $\text{v}=-\text{u}\log_\text{e}\text{M}+\text{C}$ where C is a constant.
View full question & answer→Question 163 Marks
A truck of mass $1000\ kg$ is pulling a trailer of mass $2000\ kg$ as shown. The retarding $($frictional$)$ force on the truck is $500N$ and that on the trailer is $1000N$. The truck engine exerts a force of $6000N$. Calculate:
- The acceleration of the truck and the trailer.
- The tension in the connecting rope.
Answer$i.$ The net force $f_1$ exerted on the trailer in the direction of $f_1=(T-1000) N$, where $T$ is tension in the connecting rope.
$\therefore \mathrm{T}-1000=2000 \mathrm{a} \ldots(\mathrm{i})$
Similarly, the net force $\mathrm{f}_2$ exerted by the engine of the truck is given by,
$\mathrm{f}_2=(6000-500-T)=1000 \mathrm{a}$
or $5500-T=1000 \mathrm{a} \ldots \text {..(ii) }$
Adding $(i)$ and $(ii)$, we have
$4500=3000 \mathrm{a}$
or $\mathrm{a}=\frac{4500}{3000}=1.5 \mathrm{~ms}^{-2}$
$ii.$ Putting the value of ' $a$ ' in $(i)$ we have,
$T=1000+2000 a$
$=1000+2000 \times 1.5$
$=4500 N$
View full question & answer→Question 173 Marks
Explain how momentum conservation explains explosion of a mass at rest.
AnswerExploding mass is initially at rest. The momentum is zero. When the explosion takes place, since there is no external force acting, the momentum associated with all parts should vectorially add up to zero. The velocity associated with each part is due to the conversion of internal energy into kinetic energy.
View full question & answer→Question 183 Marks
Two mutually perpendicular forces of $8N$ and $6N$ acts on the same body of mass $10\ kg$. Calculate
- Net force acting on the body,
- Magnitude of the acceleration of the body,
- Direction of acceleration of the body.
AnswerHere $\text{m}=10\text{kg},\text{F}_1=8\text{N},$
$\text{F}_2=6\text{N},\theta=90^\circ$
- Net force action on the body is,
$\text{F}=\big[\text{F}^2_1+\text{F}^2_2+2\text{F}_1\text{F}_2\cos\theta\big]^\frac{1}{2}$
$=\big[\text{F}^2_1+\text{F}^2_2\big]^\frac{1}{2}$
$[\because\cos\theta=\cos90^\circ=02]$
$=\big[8^2+6^2\big]^\frac{1}{2}=10$ or $\text{F}=10\text{N}$
- Now, $\text{F}=\text{m},\text{a}$
$\therefore\text{a}=\frac{\text{F}}{\text{m}}=\frac{10\text{N}}{10\text{kg}}=1\text{ms}^{-2}$
- Let $\alpha$ be the angle made by resultant force $(F)$ or the acceleration with $F_1$
$\therefore\tan\alpha=\frac{\text{F}_2}{\text{F}_1}=\frac{6}{8}=\frac{3}{4}=0.7500$
$\alpha=36^\circ53'$
$\therefore$ Magnitude of acceleration = $1ms^{-2}$ and it makes an angle of $36^\circ 53'$ with $8N$ force. View full question & answer→Question 193 Marks
A body placed on a rough inclined plane just begins to slide, when the slope of the plane equal to 1 in 4. Calculate the coefficient of friction.
AnswerThe slope of the plane equal to 1 in 4 implies that if BC = 4 and AB = 1. Suppose that the plane is inclined at angle 0 with the horizontal AC. From the relation between the coefficient of friction and angle of repose, we have, 
$\mu=\tan\theta$ [here, $\theta$ is angle of repose] $\mu=\frac{\text{AB}}{\text{AC}}=\frac{\text{AB}}{\sqrt{\text{BC}^2-\text{AB}^2}}$ $=\frac{1}{\sqrt{4^2-1^2}}=\frac{1}{\sqrt{16-1}}$ $\mu=\frac{1}{\sqrt{15}}$ $\mu=0.258$ View full question & answer→Question 203 Marks
A cricket bowler releases the ball in two different ways.
- Giving it only horizontal velocity, and
- Giving it horizontal velocity and a small downward velocity. The speed $v_s$ at the time of release is the same. Both are released at a height H from the ground. Which one will have greater speed when the ball hits the ground? Neglect air resistance.
AnswerFor (a) $\frac{1}{2}\text{v}_\text{z}^2=\text{gH}\Rightarrow\text{v}_\text{z}\sqrt{2\text{gH}}$ Speed at ground $=\sqrt{\text{v}_\text{s}^2+\text{v}_\text{z}^2}=\sqrt{\text{v}_\text{s}^2+2\text{gH}}$ For (b) also $[\frac{1}{2}\text{mv}_\text{s}^2+\text{mg}\text{H}]$ is the total energy of the ball when it hits the ground So the speed would be the same for both (a) and (b).
View full question & answer→Question 213 Marks
A body of mass m is placed on the floor of a lift. Find its apparent weight when the lift is:
- Moving upward with uniform acceleration.
- Moving downward with uniform acceleration.
- Moving upward with constant speed.
AnswerMass of body is ‘m placed on the floor of a lift.
- When the lift is moving upward with uniform acceleration: Suppose uniform upward acceleration of the person in the lift = a
$\therefore$ Net upward force on the person f = ma
$f = R_1 - mg$ (from fig.)
$\Rightarrow R_1 = mg + f$
$\Rightarrow R_1 = mg + ma$
$\Rightarrow R_1 = m(g + a)$
- When the lift is moving downward with uniform acceleration: Suppose uniform downward acceleration of the person in the lift = a
$\therefore$ Net downward force on the person, f = ma
From Figure it is clear that
$f = mg - R_2$
$R_2 = mg - f = mg - ma$
$R_2 = m(g - a)$
$Thus, R_2 < mg$
Apparent weight of the person becomes less than the actual weight.
- When the lift is moving upward with constant speed: Acceleration of the person = 0
$\therefore$ Net force on the person $f = 0$
i.e., $R - mg = 0$
$\Rightarrow R = mg.$ View full question & answer→Question 223 Marks
A particle of mass $0.2kg$ attached to a massless string is moving in a vertical circle of radius $1.2m$. It is imparted a speed of $8ms^{-1}$ at the lowest point of its circular path. Does the particle complete the vertical circle? What is the change is tension in the string when the particle moves from the position where the string is vertical to the position where the string is horizontal?
AnswerHere $m = 0.2kg r = 1.2m V_1 = 8ms^{-1} $
$\therefore$ Min speed required to the particle at the lowest point of the circular path $=\sqrt{\text{5gr}}=\sqrt{5\times10\times1.2}$
$=\sqrt{60}=7.7\text{ms}^{-1}$ As the speed given to the particle at the lowest point of the circular path is $8ms^{-1}$ which is more than $7.7ms^{-1}$, therefore, the particlewill certainly complete the circle. Now the change of tension in the string at the lowest point and the point where the string is horizontal = 3mg = 3 × 0.2 × 10 = 6N
View full question & answer→Question 233 Marks
Weights of $50g$ and $40g$ are connected by a stringpassing over a smooth pulley. If the system travels $2.18m$ in the first $2$ seconds, find the value of g.
AnswerDistance travelled by the system, $S = 2.18m, t_1 = 2s \text{m}_1=\frac{50}{1000}=0.05\text{kg}$
$\text{m}_2=\frac{40}{1000}=0.04\text{kg}$ Initial velocity, u = 0 Using, $\text{S}=\text{ut}+\frac{1}{2}\text{at}^2$
$\Rightarrow\ 2.18=\frac{1}{2}\times\text{a}\times4$
$\text{or a}=1.09\text{ms}^{-2}$
$\text{Also, a}=\Big(\frac{\text{m}_1-\text{m}_2}{\text{m}_1+\text{m}_2}\Big)\text{g}$
$1.09=\Big(\frac{0.05-0.04}{0.05+0.04}\Big)\times\text{g}$
$\text{or g}=\frac{1.09\times0.09}{0.01}=9.81\text{ms}^{-2}$
View full question & answer→Question 243 Marks
A man of mass $70kg$ stands on a weighing scale in a lift which is:
- Moving upwards with a uniform speed of $10m/ s$.
- Moving down with a uniform acceleration of $5m/ s^2$.
- Freely falling under gravity.
What would be reading on the scale in each case? Answer$\mathrm{m}=70 \mathrm{~kg}$
a. Uniform speed of $10 \mathrm{~ms}^{-1}$.
Since $a=0$, weight $=m g=700 \mathrm{~N}$.
b. For uniform downward acceleration,
$\mathrm{N}=\mathrm{mg}-\mathrm{ma}$.
$\therefore$ Weight $=\mathrm{mg}-\mathrm{ma}=700-70 \times 5=350 \mathrm{~N}$.
c. Force free fall $\mathrm{N}=0$.
$\therefore$ Weight felt is zero.
View full question & answer→Question 253 Marks
A constant retarding force of $50N$ is applied to a body of mass $20kg$ moving initially with a speed of $15ms^{-1}$. How long does the body take to stop?
AnswerRetarding force, F = –50N Mass of the body, m = 20kg Initial velocity of the body, u = 15m/s Final velocity of the body, v = 0 Using Newton’s second law of motion, the acceleration (a) produced in the body can be calculated as: F = ma –50 = 20 × a $\therefore\text{a}=\frac{-50}{20}=-2.5\text{ms}^{-2}$ Using the first equation of motion, the time (t) taken by the body to come to rest can be calculated as: v = u + at $\therefore\text{t}=\frac{-\text{u}}{\text{a}}=\frac{-15}{-2.5}=6\text{s}$
View full question & answer→Question 263 Marks
A man of mass $70kg$ stands on a weighing scale in a lift which is moving. Downwards with a uniform acceleration of $5ms^{-2}$.
AnswerMass of the man, $\mathrm{m}=70 \mathrm{~kg}$ Acceleration, $a=5 \mathrm{~m} / \mathrm{s}^2$ downward Using Newton's second law of motion, we can write the equation of motion as: $R+m g=m a R=m(g-a)=70(10-5)=70 \times 5=350 \mathrm{~N}$
$\therefore$ Reading on the weighing scale $=350\text{g}=\frac{350}{10}=35\text{kg}$
View full question & answer→Question 273 Marks
A truck tows a trailer of mass $1200kg$ at a speed of $10ms^{-1} $ on a level road. The tension in the coupling is $1000N$. What is the power extended on the trailer? Find the tension in the coupling when the truck ascends a road having an inclination of $1$ in $6$. Assume that the frictional resistance on the inclined plane is the same as that on the level road. 
AnswerForce applied by the truck = 1000N Power or work done per second$=\frac{\text{F}\times\text{S}}{\text{t}}=1000\times\frac{10}{1}=10^4\text{W}$
When the truck ascends a road having an inclination of 1 in 6, i.e., if $OB = 1, AB = 6$,
then it has to apply not only a forward force of $1000N$ to overcome friction, but also
will have to overcome downward component of weight, i.e., $\text{mg }\sin\theta.$
$\therefore$ Tension in the coupling, P = forward force $=1000+\text{mg}\sin\theta$
$=1000+1200\times9.8\times\frac{1}{6}$ or $P = 2960N$
Thus, the required tension in the coupling is $2960N.$ View full question & answer→Question 283 Marks
A person of mass m is standing in a lift. Find his apparent weight when the lift is:
- Moving upward with uniform acceleration a.
- Moving downward with uniform acceleration a (< g).
- Falls freely. (g is the acceleration due to gravity).
Answer
- Apparent weight = mg + ma
- Apparent weight = mg - ma
- Apparent weight = 0
View full question & answer→Question 293 Marks
State the laws of limiting friction. Hence define coefficient of friction.
AnswerThe laws of limiting friction are as follows:
- The value of limiting friction depends on the nature of the two surfaces in contact and on the state of their smoothness.
- The force of friction acts tangential to the surfaces in contact in a direction opposite to the direction of relative motion.
- The value of limiting friction is directly proportional to the normal reaction between the two given surfaces.
- For any two given surfaces and for a given value of normal reaction the force of limiting friction is independent of the shape and surface area of surfaces in contact. Coefficient of limiting friction for two given surfaces in contact is defined as the ratio of the force of limiting friction $f_1$ between them and the force of normal reaction N.
$\therefore\mu_\text{l}=\frac{\int_\text{l}}{\text{N}}$. View full question & answer→Question 303 Marks
A bob of mass $0.1kg$ hung from the ceiling of a room by a string $2m$ long is set into oscillation. The speed of the bob at its mean position is $1ms^{-1}$ . What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position.
Answer
- Vertically downward: At the extreme position, the velocity of the bob becomes zero. If the string is cut at this moment, then the bob will fall vertically on the ground.
- Parabolic path: At the mean position, the velocity of the bob is 1m/s. The direction of this velocity is tangential to the arc formed by the oscillating bob. If the bob is cut at the mean position, then it will trace a projectile path having the horizontal component of velocity only. Hence, it will follow a parabolic path.
View full question & answer→Question 313 Marks
- Explain the term impulse. Show that impulse of a variable force is equal to the area enclosed by the force-time curve.
- Two masses $8kg$ and $12kg$ are connected at the two ends of a light inextensible string that passes over a frictionless pulley. Find the acceleration of the masses and tension in the string, when the masses are released.
Answer
- The product of the force and the time interval on which it acts or change in momentum is called impulse.
$\vec{\text{I}}-\vec{\text{F}}_\text{av}\times\text{t}=\text{dp}=\text{F.dt}=$ Area of shaded portion
Net impulse $=\int\limits_{\text{t}_1}^{\text{t}_2}\vec{\text{F}}\times\text{dt}=$ area under graph ABC and time axis.
- Consider two masses $m_1$ and $m_2$ are connected to the ends of an inextensible string passing over a smooth frictionless pulley.
Let T be the tension in the string which is uniform throughout.
The heavier mass $m_1$ moves downward with an acceleration a.
The resultant downward force while considering the heavier mass m, is given by
$m_1g - T = m_1a ...(i)$
The resultant upward force while considering the lighter mass $m_2$ is given by
$T - m_2g = m_2a ...(ii)$
Adding (i) and (ii), we have
$(m_1 - m_2)g = (m_1 + m_2)a$
$\Rightarrow\ \text{a}=\Big(\frac{\text{m}_1-\text{m}_2}{\text{m}_1+\text{m}_2}\Big)\text{g}\dots(\text{iii})$
Putting the value of a in (i) and simplifying we get
$\text{T}=\frac{2\text{m}_1\text{m}_2}{\text{m}_1+\text{m}_2}\text{g}\dots(\text{iv})$
Substituting the given values, we have
$\Rightarrow\ \text{a}=\Big(\frac{12-8}{12+8}\Big)\times9.8=1.96\text{ms}^{-2}$
$\text{T}=\frac{2\times12\times8\times9.8}{12+8}=\frac{1881.6}{20}$
$=64.08\text{N}$ View full question & answer→Question 323 Marks
Discuss graphical method for the measurement of impulse in the following case:
- When constant force acts on the body.
- When a variable force acts on the body.
Answer
-
$F_1$ for curve (a) is greater than $F_2$ for curve (b). The time $t_2$ for which $F_2$ acts is greater for curve (b) than time $t_1$ in case of curve (a).
$F_1 \times t_1=F_2 \times t_2$
-
Impulse of variable force
$=\int\limits^{\text{t}_2}_{\text{t}_1}\vec{\text{F}}\text{dt}=\text{aera of }\Delta\text{BCA}$
View full question & answer→Question 333 Marks
A block placed on a rough horizontal surface is pulled by a horizontal force F. Let f be the force applied by the rough surface on the block. Plot a graph of f versus F.
AnswerWhen a small force $F_1$ is applied on a heavier box, it does not move At this state force of friction $f_1$ is equal to $F_1$. On increasing force box does not move till $F = F_s$ the maximum static frictional or limiting force. Its corresponding frictional force $f_s$ on Y-axis.
After force $F_s$, the frictional force decrease i.e., less force $\text{F}_\text{k}<\text{F}_\text{s}$ is applied on body and it starts to move with less friction $\text{f}_\text{k}<\text{f}_\text{s}$A = limiting frictional force and at B = kinetic frictional force. View full question & answer→Question 343 Marks
A body of mass $60kg$ is dragged with just enough force to start moving on a rough surface with coefficients of static and kinetic friction $0.5$ and $0.4$ respectively on applying the same force. What is the acceleration?
AnswerStatic friction $\text{f}_\text{ms}=\mu_\text{s}\text{R}$ Kinetic friction $\text{f}_\text{k}=\mu_\text{k}\text{R}$ When the body is in motion and tle applied force is $f_{ms}$, net force acting on the body i.e., $\text{F}=\text{f}_\text{ms}-\text{f}_\text{k}=\mu_\text{s}\text{R}-\mu_\text{k}\text{R}$
$=(\mu_\text{s}-\mu_\text{k})\text{mg}$
$\text{a}=\frac{\text{F}}{\text{m}}=\frac{(\mu_\text{s}-\mu_\text{k})}{\text{m}}\text{mg}$
$=(0.5-0.4)\times9.8$
$=0.98\text{ m/s}^2$
View full question & answer→Question 353 Marks
A body of mass 2 kg is being dragged with a uniform velocity of $2 \mathrm{~ms}^{-1}$ on a rough horizontal plane. The coefficient of friction between the body and the surface is 0.2 . Calculate the amount of heat generated per second.
Take $\mathrm{g}=9.8 \mathrm{~ms}^{-2}$ and $\mathrm{J}=4.2 \mathrm{~J} / \mathrm{cal}^{-1}$.
AnswerGiven, $\text{m}=2\text{kg},\text{u}=2\text{ms}^{-1},\mu=0.2$ Force of friction, $\text{F}=\mu\text{R}$
$\text{F}=\mu\text{mg}$
$[\because\text{R}=\text{mg}]$
$\text{F}=0.2\times2\times9.8$
$\text{F}=3.92\text{N}$ Distance moved per second s = ut, $\text{s}=2\times1 = 2$ Work done by friction per second, W = Fs $\text{W}=3.92\times2=7.84\text{J}$ Heat produced, $\text{H}=\frac{\text{W}}{\text{J}}\Rightarrow\text{H}=\frac{7.84}{4.2}$
$\text{h}=1.87\text{cal}.$
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A bomb at rest explodes into three parts of the same mass. The moments of the two parts is $-2 p_{\mathrm{i}}$ and $\mathrm{p}_{\mathrm{j}}$. What will be the momentum of the third part?
AnswerSince initial momentum is zero, the final momentum should also be zero. Let the momentum of third part be p_3. $\therefore\ -2\text{p}_\text{i}+\text{p}_\text{j}+\text{p}_3=0$
$\Rightarrow\ \text{P}_3=2\text{p}_\text{i}-\text{p}_\text{j}$
$\Rightarrow\ |\text{p}_3|=\text{p}\sqrt{2^2+(-1)^2}$
$=\sqrt{5}\text{p}$and is directed at an angle $\theta=\tan^{-1}\Big(\frac{-1}{2}\Big)$ with the x-axis.
View full question & answer→Question 373 Marks
A person in an elevator accelerating upwards with an acceleration of $2 \mathrm{~m} \mathrm{~s}^{-2}$, tosses a coin vertically upwards with a speed of $20 \mathrm{~m} \mathrm{~s}^1$. After how much time will the coin fall back into his hand? ( $\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2}$ )
AnswerUpward acceleration of elevator $(\mathrm{a})=2 \mathrm{~m} / \mathrm{s}^2$ Acceleration due to gravity $(\mathrm{g})=10 \mathrm{~ms}^{-2} \therefore$ Net effective acceleration $\mathrm{a}^{\prime}$
$=(a+g)=(2+10) a^{\prime}=12 \mathrm{~ms}$ Consider the effective motion of coin $v=0 t=$ time of coin to achieve maximum height $\mathrm{U}=20 \mathrm{~ms}^{-1} \mathrm{a}^{\prime}=12 \mathrm{~ms}^{-2} \therefore \mathrm{v}=\mathrm{u}+$ at here $\mathrm{a}=\mathrm{a}^{\prime} 0=20-12 \mathrm{t}$ (upward motion) $\mathrm{t}=\frac{20}{12} \mathrm{~s}=\frac{5}{3} \mathrm{~s}$ Time of ascent is equal to the time of decent.
$\therefore$ Total time to return in hand after achieving maximum height $\frac{5}{3}+\frac{5}{3}=\frac{10}{3}=3 \frac{1}{3} \mathrm{sec}$.
View full question & answer→Question 383 Marks
A block is gently placed at the top of an inclined plane $6.4m$ long. Find the time taken by the block to slide down to the bottom of plane. The plane makes an angle $30°$ with the horizontal. Coefficient of friction between the block and the plane is $0.2$. Take $g = 10m/sec^2$.
Answer$\text{a}=\text{g}(\sin\theta-\mu_\text{k}\cos\theta)$$=9.8(\sin30^\circ-0.2\cos30^\circ)=3.20\text{m/ s}^2$
$\text{S}=\text{ut}+\frac{1}{2}\text{at}^2$
$6.4=0+\frac{1}{2}\times3.2\times\text{t}^2$
$\text{t}=2\text{ sec.}$
View full question & answer→Question 393 Marks
A string of length L and mass M is lying on a horizontal table. A force F is applied at one of its end. What is the tension in string at a distance x from the end at which force is applied?
AnswerIf $\rho$ is the mass per unit length of the string, mass of entire string $\text{m}=\text{L}\rho.$ Acceleration produced in the string, $\text{a}=\frac{\text{F}}{\text{m}}=\frac{\text{F}}{\text{L}\rho}$ Force acting on the length (L - x) of the string = mass of (L - x) length of string × acceleration (a) $=(\text{L}-\text{x})\frac{\rho\times\text{F}}{\text{L}\rho}=\frac{\text{F}(\text{L}-\text{x})}{\text{L}}$
View full question & answer→Question 403 Marks
In a circus, the diameter of globe of death is 30m. From what minimum height must a cyclist start in order to roll down the inclined and go round the globe successfully?
AnswerDiameter of globe = 30m Radius of globe, r = 15m Let 'h' be the minimum height from which the cyclist after rolling down an incline will acquire velocity $=\sqrt{2\text{gh}}$ For looping the loop, the minimum velocity at the lowest point should be $\sqrt{5\text{gr}}$. $\therefore\sqrt{5\text{gr}}=\sqrt{\text{2gh}}$ or $\text{h}=\frac{5\text{r}}{2}=\frac{5\times15}{2}=37.5\text{m}.$
View full question & answer→Question 413 Marks
A $20kg$ box is gently placed on a rough inclined plane of inclination $30°$ with horizontal. The coefficient of sliding friction between the box and the plane is $0.4$. Find the acceleration of the box down the incline.
Answer
In solving inclined plane problems, the x and y directions along which the forces are to be considered, may be taken as shown.The components of weight of the box are:
- mg sin a acting down the plane.
- mg cos a acting perpendicular to the plane.
$\text{N}=\text{mg}\cos\alpha$
$\text{mg}\sin\alpha-\mu\text{N}=\text{ma}$
$\text{mg}\sin\alpha-\mu\text{mg}\cos\alpha=\text{ma}$
$\text{a}=\text{g}\sin\alpha-\mu\text{g}\cos\alpha$
$=\text{g}(\sin\alpha-\mu\cos\alpha)$
$=9.8\Big(\frac{1}{2}-0.4\times\frac{\sqrt{3}}{2}\Big)$
$=4.9\times0.3072=1.505\text{m/ }{\text{s}}^2$ The box accelerates down the plane at $1.505m/ s^2$. View full question & answer→Question 423 Marks
A woman throws an object of mass $500g$ with a speed of $25m s^1$. If the object hits a wall and rebounds with half the original speed, what is the change in momentum of the object?
Answer$\text{m}=0.5\text{kg}\ \ \text{u}=+25\text{ms}^{-1}\text{(Forword)}$
$\text{v}=\frac{-25}{2}\text{ms}^{-1}$ (as backward) $\therefore\Delta\text{p}=\text{m}(\text{v}-\text{u})=0.5\Big[\frac{-25}{2}-25\Big]$
$=0.5[-12.5-25] = 0.5\times(-37.5)$
$\Delta\text{p}=-18.75\text{kg}\ \text{ms}^{-1}$ or N-s Hence, the $\Delta\text{p}$ or $\frac{\Delta\text{p}}{\Delta\text{t}}$ or force is opposite to the initial velocity of ball.
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A stone of mass $0.2kg$ is tied to one end of a string of length $80cm$. Holding the other end, the stone is whiled into a vertical circle. What is the minimum speed of the stone at the lowest point so that it just completes the circle. What is the tension in the string at the lowest point of the circular path? ($g = 10ms^{-2}$)
AnswerHere m = 0.2kg r = 80cm = 0.8m
$\therefore V_{min}$ at the lowest point so that the stone is just able to complete the circle. $=\sqrt{\text{5gr}}=\sqrt{5\times10\times0.8}$
$=\sqrt{40}=6.32\text{ms}^{-1}$ Now tension in the string at the lower point of the circular path, $=\frac{\text{mV}^2_{\text{min}}}{\text{r}}+\text{mg}$
$\text{5mg}+\text{mg}\Big[\because\frac{\text{mV}^2_\text{min}}{\text{r}}=\text{5}\text{mg}\Big]$
$=6\text{m}=6\times0.2\times10=12\text{N}.$
View full question & answer→Question 443 Marks
Prove that the coefficient of static friction is “tangent” of the angle of repose.
AnswerThe angle of repose is defined as the maximum inclination of the plane for which the mass kept over it can stay at rest. Let $\phi$ be the angle of repose. Then,
$\text{N}=\text{mg}\cos\phi,\text{ F}_\text{f}=\text{mg}\sin\phi$
Also, $\text{F}_\text{f}=\mu\text{N}$
$\therefore\ \mu\text{ mg}\cos\phi=\text{mg}\sin\phi$
i.e., $\mu=\tan\phi.$
Since the body is at rest, the friction is static friction and the coefficient is for static case. View full question & answer→Question 453 Marks
A cricket ball of mass $150g$ is moving with a velocity of $12ms^{-1}$ and is hit by a bat so that the ball is turned back with a velocity of $20ms^{-1}$. The force of the blow acts for $0.01s$. Find the average force exerted on the ball by the bat.
AnswerThe impulse of the force exerted by the bat is given by the change in the momentum of the ball. Now, Initial momentum of the ball $=\frac{150}{1000}\times12\text{kg/ }\text{ms}^{-1}$ Final momentum of the ball $=-\frac{150}{1000}\times20\text{kg/ ms}^{-1}=-3.0\text{kg/ ms}^{-1}$ Change in the momentum of the ball $=\big[1.8-(-3.0)\big]\text{kg/ ms}^{-1}$
$=4.8\text{kg/ ms}^{-1}$ This equals the impulse of the force exerted by the bat. Since Impulse = force \times time We have, Average force exerted $=\frac{\text{Impulse}}{\text{time}}=\frac{4.8\text{kg/ ms}^{-1}}{0.01\text{s}}$
$=480\text{kg/ ms}^{-2}=480\text{N}.$
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A man of mass $70kg$ stands on a weighing scale in a lift which is moving. Upwards with a uniform speed of $10 ms^{-1}$.
AnswerMass of the man, $m=70 \mathrm{~kg}$ Acceleration, $a=0$ Using Newton's second law of motion, we can write the equation of motion as: $\mathrm{R}-\mathrm{mg}=\mathrm{ma}$ Where, ma is the net force acting on the man. As the lift is moving at a uniform speed, acceleration $\mathrm{a}=0 \therefore \mathrm{R}=\mathrm{mg}=70 \times 10=700 \mathrm{~N}$
$\therefore$ Reading on the weighing scale $=\frac{700}{\mathrm{~g}}=\frac{700}{10}=70 \mathrm{~kg}$
View full question & answer→Question 473 Marks
A ball moving with a momentum of $15kg ms^{-1}$ strikes against the wall at an angle of $30°$ and is reflected with the same momentum at the same angle. Calculate impulse.
AnswerInitial momentum $\vec{\text{P}}=15\text{kg m/s}$
Resolving it into two components$\text{p}_\text{y}=\text{p}\cos30^\circ,\text{ p}_\text{x}=\text{p}\sin30^\circ$
Final momentum $\overrightarrow{\text{p}'}=15\text{kg m/s}$ $\therefore\ \vec{\text{p}}=-\overrightarrow{\text{p}'}$ Resolving $\vec{\text{p}}$ into two components $\text{p}'_\text{y}=\text{p}'\cos30^\circ,\text{p}'_\text{x}=\text{p}'\sin30^\circ$ The two x-components are in opposite direction so they cancel out.
$\therefore$ Impulse = change in momentum$=\text{p}_\text{y}+\text{p}'_\text{y}=2\times\text{p}\cos30^\circ$
$=30\times\frac{\sqrt{3}}{2}=15\sqrt{3}\text{kg m/s}$ View full question & answer→Question 483 Marks
In the given arrangement, if the points P and Q move down with a velocity u, find the velocity of M?
AnswerUsing Pythagoras theorem, $l^2 = x^2 + y^2$. Differentiating both sides, (w.r.t. time)
$2\text{l}\frac{\text{dl}}{\text{dt}}=2\text{y}\frac{\text{dy}}{\text{dt}}$ since x is constant.
$\frac{\text{dy}}{\text{dt}}=\frac{\text{l}}{\text{y}}\frac{\text{dl}}{\text{dt}}=\frac{1}{\cos\theta}.\text{u}$
$\therefore$ Velocity of M going up $=\frac{\text{u}}{\cos\theta}$ View full question & answer→Question 493 Marks
A cyclist goes round a circular track of 440 metres length in 20 seconds. Find the angle that the cycle makes with the vertical.
AnswerLength of the track = 440m $\therefore\ \text{Radius}=\frac{440}{2\pi}\text{m}$ $\text{Speed}=\frac{2\pi\text{r}}{\text{t}}=\frac{440}{20}=22\text{ ms}^{-1}$ 
$\tan\theta=\frac{\text{v}^2}{\text{rg}}$ $\therefore\ \theta=\tan^{-1}\Bigg[\frac{(22)^2}{\frac{440}{2\pi}\times\text{g}}\Bigg]=35^\circ12'$ View full question & answer→Question 503 Marks
Two masses $m_1$ and $m_2$ are connected to the ends of a string passing over a pulley. Find the tension and acceleration associated.
AnswerThe pulley is frictionless, massless and fixed. The free body diagram for the two masses are shown along with the equation.$\text{m}_2\text{g}-\text{T}=\text{m}_2\text{a},$
$\text{T}-\text{m}_1\text{g}=\text{m}_1\text{a}$
Solve the equations to get,
$\text{a}=\frac{(\text{m}_2-\text{m}_1)\text{g}}{(\text{m}_2+\text{m}_1)}$ and $\text{T}=\frac{2\text{m}_1\text{m}_2\text{g}}{(\text{m}_1+\text{m}_2)}.$ View full question & answer→