Question 12 Marks
Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg , (Neglect air resistance throughout.) Just after it is dropped from the window of a train accelerating with $1 \mathrm{~ms}^{-2}$.
Answer1 N ; vertically downward It is given that the train is accelerating at the rate of $1 \mathrm{~m} / \mathrm{s}^2$. Therefore, the net force acting on the stone, $\mathrm{F}^{\prime}=\mathrm{ma}=0.1 \times 1=0.1 \mathrm{~N}$ This force is acting in the horizontal direction. Now, when the stone is dropped, the horizontal force F', stops acting on the stone. This is because of the fact that the force acting on a body at an instant depends on the situation at that instant and not on earlier situations. Therefore, the net force acting on the stone is given only by acceleration due to gravity. $\mathrm{F}=\mathrm{mg}=1 \mathrm{~N}$ This force acts vertically downward.
View full question & answer→Question 22 Marks
Give the magnitude and direction of the net force acting on a stone of mass 0.1kg, (Neglect air resistance throughout.) Just after it is dropped from the window of a train running at a constant velocity of 36km/h.
Answer1N; vertically downward The train is moving with a constant velocity. Hence, its acceleration is zero in the direction of its motion, i.e., in the horizontal direction. Hence, no force is acting on the stone in the horizontal direction. The net force acting on the stone is because of acceleration due to gravity and it always acts vertically downward. The magnitude of this force is 1N.
View full question & answer→Question 32 Marks
Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg , (Neglect air resistance throughout.) Lying on the floor of a train which is accelerating with $1 \mathrm{~ms}^{-2}$, the stone being at rest relative to the train.
Answer0.1 N ; in the direction of motion of the train, The weight of the stone is balanced by the normal reaction of the floor. The only acceleration is provided by the horizontal motion of the train. Acceleration of the train, $a=0.1 \mathrm{~m} / \mathrm{s}^2$ The net force acting on the stone will be in the direction of motion of the train. Its magnitude is given by: $\mathrm{F}=\mathrm{ma}=0.1 \times 1=$ 0.1 N
View full question & answer→Question 42 Marks
Give the magnitude and direction of the net force acting on a stone of mass 0.1kg, (Neglect air resistance throughout.) Just after it is dropped from the window of a stationary train.
Answer1 N ; vertically downward Mass of the stone, $m=0.1 \mathrm{~kg}$ Acceleration of the stone, $a=g=10 \mathrm{~m} / \mathrm{s}^2$ As per Newton's second law of motion, the net force acting on the stone, $\mathrm{F}=\mathrm{ma}=\mathrm{mg}=0.1 \times 10=1 \mathrm{~N}$ Acceleration due to gravity always acts in the downward direction.
View full question & answer→Question 52 Marks
Is a 'single isolated force' possible in nature?
AnswerA single isolated force is not possible. This follows from Newton's third law of motion, according to which to every action, there is an equal and opposite reaction. So, the forces must always exist in pairs. When we talk of a single force, we are considering only one aspect of mutual interaction.
View full question & answer→Question 62 Marks
A tennis ball of mass 'm' strikes a massive wall with a velocity V and traces back the same path. Calculate the change in momentum of the ball due to the wall.
AnswerLet $\mathrm{m}_1=$ mass of ball $\mathrm{v}_1=$ velocity of ball $=v \mathrm{~m}_2=$ mass of wall $\mathrm{v}_2=$ velocity of wall $=0 \mathrm{~m}_2 \gg m_1$ ( $\mathrm{m}_1$ can be neglected) $\therefore$ Change in momentum of ball $=$ final momentum - initial momentum $=\mathrm{m}[\mathrm{v}-(-\mathrm{v})]=2 \mathrm{mv}$.
View full question & answer→Question 72 Marks
A monkey is ascending a branch with constant acceleration. If the breaking strength is 160% of the monkey's weight, what is the maximum acceleration permitted for the monkey?
AnswerBreaking strength $=\frac{160}{100}\text{mg},$ where m is the mass of the monkey. While ascending the apparent weight w = m(g + a).For safety, $\text{m(g + a)}\leq\frac{160}{100}\text{mg}$
$\therefore\ \text{a}\leq\Big(\frac{160}{100}-1\Big)\text{g,}$
$\text{i.e., a}\leq0.6\text{m/ s}^2.$
View full question & answer→Question 82 Marks
It is easier to pull a lawn roller than to push it. Why?
AnswerWhen we pull the lawn roller, the vertical component of the applied pull acts opposite to the weight of the roller and it reduces its effective weight. On the other hand, when the lawn roller is pushed, the vertical component of the applied push adds to the weight of the roller and therefore its effective weight increases. As the effective weight is lesser, when the lawn roller is pulled, it is easier to pull the lawn roller than to push it.
View full question & answer→Question 92 Marks
A woman throws an object of mass 500g with a speed of $25m s^1$. What is the impulse imparted to the object?
AnswerMass of object m = 500 g = 0.5kg u = 0, v = 25m/ s Impulse $\vec{\text{F}}.\text{dt}=\frac{\text{d}\vec{\text{p}}}{\text{dt}}=\text{d}\vec{\text{p}}=\text{m}\vec{\text{v}}-\text{m}\vec{\text{u}}$ $\text{I}=\Delta\vec{\text{p}}=\text{m}(\vec{\text{v}}-\vec{\text{u}})=0.5(25.5)\text{N}-\text{s}$
View full question & answer→Question 102 Marks
A person of mass 50 kg stands on a weighing scale on a lift. If the lift is descending with a downward acceleration of $9 \mathrm{~m} \mathrm{~s}^{-2}$, what would be the reading of the weighing scale? $\left(\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2}\right)$
AnswerWhen lift is descending with acceleration a, the apparent weight decreases on weighing scale $\therefore\text{W}'=\text{R}=(\text{mg}-\text{ma})=\text{m}(\text{g}-\text{a})$ Apparent weight due to reaction force by the lift on weighing scale. $\therefore\text{W}'=50(10-9)=50\text{N}$ Reading of weighing scale $=\frac{\text{R}}{\text{g}}=\frac{50}{10}=5\text{kg}$
View full question & answer→Question 112 Marks
Two masses M and m are connected at the two ends of an inextensible, light string. The string passes over a smooth frictionless pulley. Calculate the acceleration of the masses and the tension in the string. M is heavier than m.
AnswerThe pulley is frictionless, massless and fixed. The free body diagram for the two masses are shown along with the equation. 
Mg - T = Ma T - mg = ma Solving the equation, we get $\text{a}=\frac{(\text{M}-\text{m})\text{g}}{\text{M}+\text{m}}$ and $\text{T}=\frac{(2\text{Mm})\text{g}}{\text{M}+\text{m}}$ View full question & answer→Question 122 Marks
Write the three laws of motion.
Answer
- A body at rest or uniform motion will continue to maintain the status, till an unbalanced force acts on it.
- The rate of change of momentum is a measure of the force acting on the body.
Alternate Answer
The total unbalanced external force acting on a body is the product of its mass and acceleration.
- For every action there exists an equal and opposite reaction.
View full question & answer→Question 132 Marks
Is friction a necessary evil? Justify.
AnswerFriction opposes the motion and brings wear and tear. But without friction one cannot stop any body.
View full question & answer→Question 142 Marks
A aircraft executes horizontal loop of radius 1.00km with a steady speed of 900km/ h. Compare its centripetal acceleration with the acceleration due to gravity (g).
AnswerRadius of the loop, $\mathrm{r}=1 \mathrm{~km}=1000 \mathrm{~m}$ Speed of the aircraft, $\mathrm{v}=900 \mathrm{~km} / \mathrm{h}=900 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}=250 \mathrm{~m} / \mathrm{s}$ Centripetal acceleration, $\mathrm{a}_{\mathrm{c}}=\frac{\mathrm{v}^2}{\mathrm{r}}=\frac{62500}{1000}=62.5 \mathrm{~m} /$ sAcceleration due to gravity, $\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2$
$\therefore \frac{a_c}{g}=\frac{62.5}{9.8}=6.38 \mathrm{~g} \simeq 6.4 \mathrm{~g}$
View full question & answer→Question 152 Marks
A block of mass M is held against a rough vertical wall by pressing it with a finger. If the coefficient of friction between the block and the wall is $\mu$ and the acceleration due to gravity is g, calculate the minimum force required to be applied by the finger to hold the block against the wall?
AnswerLet F force is applied by the finger on a body of mass M to hold rest against the wall. Under the balanced condition 
$\text{F}=\text{N} $ and $\text{f}=\text{Mg}$$\Rightarrow\mu\text{F}=\text{Mg}$ or $\text{F}=\frac{\text{Mg}}{\mu}$ is the minimum force to hold the block against the wall at rest. View full question & answer→Question 162 Marks
A person driving a car suddenly applies the brakes on seeing a child on the road ahead. If he is not wearing seat belt, he falls forward and hits his head against the steering wheel. Why?
AnswerWhen a person applies breaks suddenly, the lower part of person slows rapidly with the car, but the upper part of driver continue to move with same speed in the same direction due to the inertia of motion and his head car hit with steering.
View full question & answer→Question 172 Marks
A mass of 2kg is suspended with thread AB Thread CD of the same type is attached to the other end of 2kg mass. Lower thread is pulled gradually, harder and harder in the downward directon so as to apply force on AB. Which of the threads will break and why?
AnswerThread AB will break. Force on CD is equal to the force (f) applied at D downward, but the force on thread AB is equal to the force F along with force due to mass 2kg downward. so the force on AB is 2kg more than applied force at D.Hence the thread AB will break up.
View full question & answer→Question 182 Marks
What are the three Newtons Laws of Motion.
AnswerNewton's First Law of Motion : Every body continues to be at rest or in motion until or unless it is compelled by an external force. Newton's Second Law of Motion : The rate of change of momentum of a body is directly proportional to the force applied and in the same direction as the force is applied. Newton's Third Law of Motion : To every action there is equal and opposite reaction.
View full question & answer→Question 192 Marks
Block A of weight 100N rests on a frictionless inclined plane of slope angle 30° A flexible cord attached to A passes over a frictonless pulley and is connected to block B of weight W. Find the weight W for which the system is in equilibrium.
AnswerMain concept used: On balanced condition i.e., no motion then no frictional force or f = 0 Explanation: During equilibrium of A or B $\text{mg}\sin30^\circ=\text{f}$ $\frac{1}{2}\text{mg}=\text{F}\ [\because\text{mg}=100\ \text{N}]$ $\therefore\text{F}=\frac{1}{2}\times100=100=50$ For B is at rest W = F =50N.
View full question & answer→Question 202 Marks
A nucleus is at rest. All of a sudden it splits into two small nuclei. What is the angle at which these two nuclei fly apart?
AnswerLet $M=$ mass of nucleus at rest, $\therefore$ Momentum of the nucleus before disintegration $=M \times 0=0$, Let $m_1$ and $m_2$ be the mass of the two smaller nuclei and $v_1$ and $v_2$ be their velocities. $\therefore$ Momentum of the nucleus after disintegration $=m_1 v_1+m_2 v_2$. According to the law of conservation of linear momentum $m_1 v_1+m_2 v_2=0 m_1 v_1=-m_2 v_2 \cdot \mathbf{O R}$ The -ve sign shows that the velocities $\mathrm{v}_1$ and $\mathrm{v}_2$ must be opposite sign i.e., the two products must be emitted in opposite direction. Thus, the angle between two nuclei is $180^{\circ}$.
View full question & answer→Question 212 Marks
A block of mass M is placed on a frictionless, inclined plane of angle $\theta,$ as shown in the figure. Determine the acceleration of the block after it is released. What is force exerted by the incline on the block? 
AnswerWhen the block is released, it will move down the incline. Let its acceleration be a. As the surface is frictionless, so the contact force will be normal to the plane. Let it be N. Here, for the block we can apply equation for motion along the plane and equation for equilibrium perpendicular to the plane. 
i.e., $\text{Mg}\sin\theta=\text{Ma}\Rightarrow\text{a}=\text{g}\sin\theta$ Also, $\text{Mg}\cos\theta-\text{N}=0$ $\Rightarrow\text{N}=\text{Mg}\cos\theta.$ View full question & answer→Question 222 Marks
A light, inextensible string as shown in figure connects two blocks of mass $M_1$ and $M_2$. A force F as shown acts upon $M_1$. Find acceleration of the system and tension in string.
AnswerHere as the string is inextensible, acceleration of two blocks will be same. Also, string is massless so tension throughout the string will be same. Contact force will be normal force only. Let acceleration of each block is a, tension in string is T and contact force between $M_1$ and surface is $N_1$ and contact force between $M_2$ and surface is $N_2$. Applying Newton's second law for the blocks; For $M_1, F - T = M_1a$ ...(i) $m_1g - N_1 = 0$ ...(ii) For $M_2, T = M_2a$ ...(iii) $M_2g - N_2 = 0$ ...(iv) Solving equations (i) and (iii) $\text{a}=\frac{\text{F}}{\text{M}_1\text{M}_2}$ and $\text{T}=\frac{\text{M}_2\text{F}}{\text{M}_1+\text{M}_2}$
View full question & answer→Question 232 Marks
Why does a gun recoil? Derive the recoil velocity of a gun.
AnswerSince no external force acts on gun, the momentum has to be conserved. So the gun recoils. If $m_g$ and $m_b$ are the mass of gun and bullet with velocity of bullet being $\mathrm{V}_{\mathrm{b}}$ then $\mathrm{m}_{\mathrm{g}} \mathrm{v}_{\mathrm{g}}+\mathrm{m}_{\mathrm{b}} \mathrm{v}_{\mathrm{b}}=0, \therefore \mathrm{v}_{\mathrm{g}}=-\frac{\mathrm{m}_{\mathrm{b}} \mathrm{v}_{\mathrm{b}}}{\mathrm{m}_{\mathrm{g}}}$-ve sign means that the gun moves opposite to the direction of the bullet.
View full question & answer→Question 242 Marks
A hammer of mass 1kg strikes on the head of a nail with a velocity of $10m s^{-1}7$.It drives the nail 1cm into a wooden block. Calculate the force applied by the hammer and the time of impact.
AnswerHere mass of hammer $M=1 \mathrm{~kg}$, when hammer strikes the nail with a velocity of $10 \mathrm{~ms}^{-1}$ and as mass of nail is extremely small, hence nail also starts moving with same velocity. Thus, for nail $\mathrm{u}=10 \mathrm{~ms}^{-1}, \mathrm{v}=0$ and $\mathrm{s}=1 \mathrm{~cm}=$ 0.01 m . Using the relation $\mathrm{v}^2-\mathrm{u}^2=2$ as, we get $(0)^2-(10)^2=2 \times \mathrm{a} \times(0.01) \Rightarrow \mathrm{a}=-\frac{10 \times 10}{2 \times 0.01}=-5 \times 10^3 \mathrm{~ms}^{-2}$ and using relation $\mathrm{v}=\mathrm{u}+$ at, we have $0=10-5 \times 10^{-3} \cdot \mathrm{t} \Rightarrow \mathrm{t}=\frac{10}{5 \times 10^3}=2 \times 10^{-3} \mathrm{~s}$ or $2 \mathrm{~ms} . \therefore$ Force exerted by the hammer on the nail $\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}=\frac{\mathrm{Mu}-0}{\Delta \mathrm{t}}=\frac{1 \times 10}{2 \times 10^{-3}}=5 \times 10^3 \mathrm{~N}$.
View full question & answer→Question 252 Marks
Name the fundamental forces of nature.
AnswerGravitational force, Nuclear force, Electro-magnetic force and weak forces.
View full question & answer→Question 262 Marks
A block is supported by a cord C from a rigid support, and another cord D is attached to the bottom of the block. If you give a sudden jerk to D, it will break. But if you pull on D steadily, C will break. Why?
AnswerString C breaks because C is stretched more than D. This is because C was already in stretched state due to large weight. When D is given a jerk, the load will receive only a small acceleration due to its large mass. Thus, C will not be further stretched but D will exceed the safe limit and break.
View full question & answer→Question 272 Marks
Derive the law of conservation of liner momentum from Newton's third law of motion.
AnswerAccording to third law, for every action there is an equal and opposite reaction. So, if $\mathrm{dP}_1$ and $\mathrm{dP}_2$ are change in momentum of two masses $\mathrm{m}_1$ and $\mathrm{m}_2$ then $\frac{\mathrm{dP}_1}{\mathrm{dt}}=-\frac{\mathrm{dP}_2}{\mathrm{dt}}$ Since, $\mathrm{F}_1=-\mathrm{F}_2 \therefore-\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{P}_1+\mathrm{P}_2\right)=0$ i.e., $\mathrm{P}_1+\mathrm{P}_2=$ constant.
View full question & answer→Question 282 Marks
Is earth an inertial frame of reference?
AnswerSince earth rotates about its own axis and also revolve round the sun, there will be acceleration associated. So earth cannot be taken as an inertial frame of reference (strictly).
View full question & answer→Question 292 Marks
Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg , (Neglect air resistance throughout.) Just after it is dropped from the window of a train accelerating with $1 \mathrm{~ms}^{-2}$.
Answer1 N ; vertically downward It is given that the train is accelerating at the rate of $1 \mathrm{~m} / \mathrm{s}^2$. Therefore, the net force acting on the stone, $\mathrm{F}^{\prime}=\mathrm{ma}=0.1 \times 1=0.1 \mathrm{~N}$ This force is acting in the horizontal direction. Now, when the stone is dropped, the horizontal force $\mathrm{F}^{\prime}$, stops acting on the stone. This is because of the fact that the force acting on a body at an instant depends on the situation at that instant and not on earlier situations. Therefore, the net force acting on the stone is given only by acceleration due to gravity. $\mathrm{F}=\mathrm{mg}=1 \mathrm{~N}$ This force acts vertically downward.
View full question & answer→Question 302 Marks
A particle of mass 200g is whirled into a vertical circle of radius 80cm using a massless string. The speed of particle when the string makes an angle $60^\circ$ with the vertical line is $1.5ms^{-1}$ What is the tension in the string is this position?
AnswerHere, m = 200g = 0.2kg r = 80cm = 0.8m $\text{v}=1.5\text{ms}^{-1}$ $\theta=60^\circ$ $\therefore$ Required tensionin the string $\text{T}=\frac{\text{mV}^2}{\text{r}}+\text{mg}\cos\theta$ $=\frac{0.2\times(1.5)^2}{0.8}+0.2\times10\times\cos60^\circ$ $\frac{2.25}{4}+0.2\times10\times\frac{1}{2}$ $=0.56+1=1.56\text{N}$
View full question & answer→Question 312 Marks
A body of mass 500 g tied to a string of length 1 m is revolved in the vertical circle with a constant speed. Find the minimum speed at which there will not be any slack on the string. Take $\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2}$.
AnswerThe tension T in the string will provide the necessary centripetal force $\frac{\text{mv}^2}{\text{r}}$ i.e., $\text{T}=\frac{\text{mv}^2}{\text{r}}$ Here, $\text{m}=500\text{g}=\frac{1}{2}\text{kg};\text{ r}=1\text{m}$ $\therefore\text{T}=\frac{1}{2}\text{v}^2\text{ N}$ There will not be slack if $\text{T}\geq$ Weight of the body. i.e., $\text{T}\geq\text{mg}$ or $\frac{1}{2}\text{v}^2\geq\frac{1}{2}\times10$ So, the minimum speed $=\sqrt{10}\text{m s}^{-1}=3.162\text{m }\text{s}^{-1}.$
View full question & answer→Question 322 Marks
Give some ways by which the friction can be reduced.
Answer
- Polishing the surfaces.
- Lubricating the surfaces.
- Streamlining of bodies.
- Use of ball bearings.
View full question & answer→Question 332 Marks
Find the minimum force required to pull a body up a rough inclined plane $(\theta,\mu).$
AnswerAs the mass has to be pulled up, the component of force of gravity and the force of friction act downward. Therefore, the minimum force required is, $(\text{mg}\sin\theta+\mu\text{ mg}\cos\theta).$
View full question & answer→Question 342 Marks
A car of mass 1000 kg moving with a speed of $30 \mathrm{~ms}^{-1}$ collides with the back of a stationary lorry of mass 9000 kg . (Fig.).Calculate the speed of the vehicles immediately after the collision if they remain jammed together.
AnswerUsing conservation of momentum, $(1000+9000)\text{v}=1000\times30+9000\times0$ or $\text{V}=\frac{1000\times30}{10000}\text{m/s}^{-1}=\text{3ms}^{-1}.$
View full question & answer→Question 352 Marks
Give the magnitude and direction of the net force acting on a stone of mass 0.1kg, (Neglect air resistance throughout.) Just after it is dropped from the window of a train running at a constant velocity of 36km/h.
Answer1N; vertically downward The train is moving with a constant velocity. Hence, its acceleration is zero in the direction of its motion, i.e., in the horizontal direction. Hence, no force is acting on the stone in the horizontal direction. The net force acting on the stone is because of acceleration due to gravity and it always acts vertically downward. The magnitude of this force is 1N.
View full question & answer→Question 362 Marks
Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg , (Neglect air resistance throughout.) Lying on the floor of a train which is accelerating with $1 \mathrm{~ms}^{-2}$, the stone being at rest relative to the train.
Answer0.1 N ; in the direction of motion of the train, The weight of the stone is balanced by the normal reaction of the floor. The only acceleration is provided by the horizontal motion of the train. Acceleration of the train, $a=0.1 \mathrm{~m} / \mathrm{s}^2$ The net force acting on the stone will be in the direction of motion of the train. Its magnitude is given by: $\mathrm{F}=\mathrm{ma}=0.1 \times 1=$ 0.1 N
View full question & answer→Question 372 Marks
Give one argument in favour of the fact that frictional force is a non conservative force.
AnswerThe direction of the frictional force is opposite to the direction of motion. When a body is moved, say from A to B and then back to A, work is required to be done both during forward and backward motion. So, the net work done in a round trip is not zero. Hence, the frictional force is a non-conservative force.
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What is the apparent weight felt by a person in an elevator, when it is accelerating $(i)$ upward by $'a\ '\ (ii)$ downward by $'a\ '\ ?$
Answer
- Apparent weight $= m(g + a).$
- Apparent weight $= m(g - a).$
View full question & answer→Question 392 Marks
Define centripetal force. A cyclist speeding at 18km/ hr on a level road takes a sharp circular turn of radius 3m without reducing the speed. The coefficient of static friction is 0.1. Will the cyclist slip while taking the turn?
AnswerThe force on the body towards the centre while it is moving is a circular path. The condition for the cyclist not to slip is $\text{V}^2\leq\mu_\text{s}\times\text{R}\times\text{g}$ $\text{V}^2\leq0.1\times3\times9.8$ $\text{V}^2=2.94\text{m}^2/\text{s}^2$ But the speed of the cyclist is $18\text{km}/\text{hr}=5\text{m}/\text{s}$ $\therefore\text{V}^2=25\text{m}^2/\text{s}^2$ $\therefore$ The condition is not obeyed $\therefore$ The cyclist will slip.
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Two bodies of mass M and m (M > m) are allowed to fall freely from the same height. If air resistance for each body be same which one reach the ground first?
AnswerThe heavier body will reach the ground earlier. Net downward force on mass, M = mg - F$\Rightarrow\ \text{a}'=\frac{\text{Mg}-\text{F}}{\text{M}}$
Similarly for mass (m),$ \text{a}'=\frac{\text{Mg}-\text{F}}{\text{m}}$
F is same on both as M > m ⇒ a > a'
View full question & answer→Question 412 Marks
A bird is sitting on the floor of a closed glass cage and the cage is in the hand of a girl. Will the girl experience any change in the weight of the cage when the bird
- Starts flying in the cage with a constant velocity;
- Flies upwards with acceleration.
- Flies downwards with acceleration?
AnswerIn a closed glass cage, air inside is bound with the cage. Therefore,
- There would be no change in weight of the cage if the bird flies with a constant velocity.
- The cage becomes heavier, when bird flies upwards with an acceleration.
- The cage appears lighter, when bird flies downwards with an acceleration.
View full question & answer→Question 422 Marks
A passenger of mass $72.2\ kg$ is riding in an elevator while standing on a platform scale. What does the scale read when the elevator cab is:
- Descending with constant velocity,
- Ascending with constant acceleration, $3.5\ m/ s$?
AnswerGiven, mass, $\mathrm{m}=72.2 \mathrm{~kg}$
Gravity acceleration, $\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2$
Scale reading $=$ apparent weight $=\mathrm{R}= ?$
i. While descending with constant velocity, $\mathrm{a}=0$
$R=m g$
$R=72.2 \times 9.8$
$R=707.56 \mathrm{~N}$
ii. While ascending with $a=3.2 \mathrm{~m} / \mathrm{s}$
$R=m(g+a)$
$R=72.2(9.8+3.2)$
$=938.6 \mathrm{~N}$
View full question & answer→Question 432 Marks
A cyclist speeding at 18km/ h on a level road takes a sharp circular turn of radius 3m without reducing the speed. The co-efficient of static friction between the tyres and road is 0.1. Will the cyclist slip while taking the turn? Explain.
AnswerThe maximum velocity with which a vehicle can go round a level curve, without skidding is, Given radius r = 3m, $\mu_\text{s}=0.1$ and $\mathrm{v}=18 \mathrm{kmh}^{-1} \text{V}_\text{max}=\sqrt{\mu_\text{s}\text{rg}}=\sqrt{0.1\times3\times10}$
= 1.732m/ s This is the maximum velocity with which cyclist can take turn. But, the velocity of the cyclist = 18km/ h $=18\times\frac{5}{18}\text{m/ s}=5\text{m/ s}$ Hence cyclists will slip.
View full question & answer→Question 442 Marks
Why is it necessary to keep the rate of fuel consumption in a rocket constant?
AnswerIf the rate of fuel consumption is not kept the same, the energy produced every moment will be different and may lead to problem of controlling large energy or accelerating the rocket.
View full question & answer→Question 452 Marks
Give the magnitude and direction of the net force acting on a stone of mass 0.1kg, (Neglect air resistance throughout.) Just after it is dropped from the window of a stationary train.
Answer1 N ; vertically downward Mass of the stone, $m=0.1 \mathrm{~kg}$ Acceleration of the stone, $a=g=10 \mathrm{~m} / \mathrm{s}^2$ As per Newton's second law of motion, the net force acting on the stone, $F=m a=m g=0.1 \times 10=1 \mathrm{~N}$ Acceleration due to gravity always acts in the downward direction.
View full question & answer→Question 462 Marks
To what angle a cyclist has to bend to negotiate a curved path? Derive an expression.
AnswerAs the cyclist leans by an angle $\theta$ with the vertical, the normal can be acting as shown. If v is the speed of the cyclist, the necessary centripetal force is provided by the component $\text{N}\sin\theta$ of normal reaction and $\text{N}\cos\theta$ provides balancing for weight. 
$\therefore\ \text{N}\sin\theta=\frac{\text{mv}^2}{\text{r}},$ $\text{N}\cos\theta=\text{mg}$ Dividing, $\tan\theta=\frac{\text{v}^2}{\text{rg}}$ $\Rightarrow\ \theta=\tan^{-1}\Big(\frac{\text{v}^2}{\text{rg}}\Big)$ View full question & answer→Question 472 Marks
Find the acceleration with a mass sliding down an inclined plane $(\theta)$ with coefficient of friction $\mu.$
AnswerAs the mass slides down, the force of friction, will act in the upward direction along the plane. The net force acting will be, $(\text{mg}\sin\theta-\mu\text{ mg}\cos\theta).$ Therefore, the acceleration is $\text{g}(\sin\theta-\mu\cos\theta).$
View full question & answer→Question 482 Marks
What is the principle behind the launch of rockets?
AnswerThe principle behind the launch of rockets is the law of conservation of momentum. The change in momentum with the burnt fuel provides momentum to the rocket.
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Air is thrown on a sail attached to a boat from an electric fan placed on the boat. Will the boat start moving?
AnswerNo, when the fan pushes the sail by air, then air also pushes the fan in the opposite direction. Since fan is placed on the boat, the vector sum of linear moments of fan and the boat is zero. The boat can move only under the force of reaction from some external agency.
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In a rotor, a hollow vertical cylinder rotates about its axis and a person rests against the inner wall. At a particular speed of the rotor, the floor below the person is removed and the person hangs resting against the wall without any floor. If the radius of the rotor is $2m$ and the coefficient of static friction between the wall and the person is $0.2.$
- Find the minimum speed at which the floor may be removed.
- What type of speciality is associated with this question? $[$take, $g = 10m/ s^2]$
AnswerThe situation is shown in figure below:
- When the floor is removed, the forces on the person are:
- Weight mg downward.
- Normal force $N$ due to the wall towards the centre.
- Frictional force $f$, parallel to the wall, upwards.
The person is moving in a circle with a uniform speed, so its acceleration is $v?$ Ir towards the centre.
Newton's law for the horizontal direction $($second law$)$ and for the vertical direction $($first law$)$ give
$N = mv^{2/ r}$ and,
$f_s = mg$
For the minimum speed, when the floor may be removed, the friction is limiting one and, so equals $\mu_\text{s}\text{N}.$
This given $\mu_\text{s}\text{N}=\text{mg}$
or $\frac{\mu_\text{s}\text{mv}^2}{\text{r}}=\text{mg} [$using Eq. $(i)]$
or $\text{v}=\sqrt{\frac{\text{rg}}{\mu_\text{s}}}=\sqrt{\frac{2\text{m}\times10\text{ms}^2}{0.2}}=10\text{ms}.$
- Speciality is that without the floor an object a body may be $0$ align with a vertical wall provided, it is set to be in circular motion $($horizontal$)$ with properly required speed.
View full question & answer→Question 512 Marks
A mass m is at rest on a rough surface. Draw the variation of the force of friction experienced with the force applied on it.
Question 522 Marks
A body is accelerating with 'a' $m/ s^2$ while its mass is increasing at the rate of 'm'. What will be the force experienced by it?
AnswerForce required to keep the acceleration is F = ma. Due to variation in mass the momentum changes. So the extra force needed is, $\text{F}=\frac{\text{dm}}{\text{dt}}.\ \vec{\text{v}},$ where $\vec{\text{v}}$ is the instantaneous velocity. The net force required is, $\text{F}=\text{m}\vec{\text{a}}+\frac{\text{dm}}{\text{dt}}\vec{\text{v}}.$
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State Impulse-momentum theorem.
AnswerThe change in momentum associated with a body can be produced by a large force for small time or a small force for large time.
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Using component of force, show that it is easier to pull a lawn roller than to push it.
Answer
Apparent weight $=\text{F }\sin\theta+\text{mg}=\text{Reaction (R)}$
$\Rightarrow\ \text{f}=\mu\text{R}$
Apparent weight $=\text{mg}-\text{F }\sin\theta$
$=\text{Reaction (R}')$
$\text{f}'=\mu\text{R}'$
As f > f'
So, pulling is easier. View full question & answer→Question 552 Marks
A spring weighing machine inside a stationary lift reads $10\ kg,$ when a man stands on it. What would happen to the scale reading, if the lift is moving upward with $(i)$ constant velocity and $(ii)$ constant acceleration?
AnswerThe apparent weight, $W = m(g + a),$ where $a$ is the acceleration with which the lift moves upwards.
- When lift moves with constant velocity, $a = 0,$
Therefore, $W = m(g + 0) = mg = 10\ kg. ($no changes$)$
- When lift moves with acceleration a in upward direction, then
$W = m(g + a) =\text{mg}\Big(1+\frac{\text{a}}{\text{g}}\Big)\geq10\ \text{kg} ($Apparent weight will increase$).$ View full question & answer→Question 562 Marks
It is known that polishing a surface beyond a certain limit increases the frictional force rather than decreasing it. Explain.
AnswerWhen the surfaces are highly polished the area of contact between them increases. A large number of atoms and molecules lying on both the surfaces start exerting strong attractive forces on each other consequently friction increases.
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A bird is sitting on the floor of a wire cage and the cage is in the hands of a boy. The bird starts flying in the cage. Will the boy experience any change in the weight of the cage?
AnswerIn a wire cage, air inside is in free contact with atmospheric air. Therefore, when the bird starts flying inside the cage the weight of bird is no more experienced. Hence the cage will appear lighter than before.
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Define force of friction. How does the use of ball bearing reduce friction?
AnswerThe force parallel to the surface of contact which opposes relative motion between the two surfaces. By using ball bearings, the wheel rolls on balls instead of sliding on axle. So the sliding friction is converted into rolling friction which is much less.
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A soda water bottle is falling freely. Will the bubbles of gas rise in the water of the bottle?
AnswerBubbles will not rise in water. This is because water in freely falling bottle is in the state of weightlessness. No reactional upward force acts on the bubbles.
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How does the lubricants help in reducing friction?
AnswerLubricants spread over the surface and fill the irregularities. The contact between the lubricant and the wheel reduces the friction.
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A monkey is descending from the branch of a tree with constant acceleration. If the breaking strength is 75% of the weight of the monkey, what is the minimum acceleration with which monkey can slide down without breaking the branch?
AnswerFor descending monkey T = mg - ma = m(g- a) As T = 75% of mg i.e., $\text{T}=\frac{3}{4}\text{mg}$ $\frac{3}{4}\text{mg}=\text{m}(\text{g}-\text{a})$ $\text{a}=\frac{\text{g}}{4}$
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What are the changes force can bring on a body?
AnswerForce can change the magnitude of speed, the direction of motion and the shape of the body.
View full question & answer→Question 632 Marks
Show that Newton's second law is the Real Law of Motion.
AnswerAccording to Newton's law, force is the rate of change of momentum. This law,
- Expresses the need for a force to accelerate a body.
- Give an estimate of force and its outcome.
So it is the basic law of force.
View full question & answer→Question 642 Marks
The motion of a particle of mass $m$ is described by $y=u t+\frac{1}{2} g t^2$. Find the force acting on the particle.
AnswerWe know
$
y=u t+\frac{1}{2} g t^2
$
Now,
$
v=\frac{ d y}{ d t}=u+g t
$
acceleration, $a=\frac{ d v}{ d t}=g$
Then the force is given by Eq. (4.5)
$
F=m a=m g
$
Thus the given equation describes the motion of a particle under acceleration due to gravity and $y$ is the position coordinate in the direction of $g$.
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A bullet of mass $0.04 kg$ moving with a speed of $90 m s ^{-1}$ enters a heavy wooden block and is stopped after a distance of $60 cm$. What is the average resistive force exerted by the block on the bullet?
AnswerThe retardation ' $a$ ' of the bullet (assumed constant) is given by
$
a=\frac{-u^2}{2 s }=\frac{-90 \times 90}{2 \times 0.6} m s ^{-2}=-6750 m s ^{-2}
$
The retarding force, by the second law of motion, is
$
=0.04 \ kg \ 6750 \ m s ^{-2}=270 N
$
The actual resistive force, and therefore, retardation of the bullet may not be uniform. The answer therefore, only indicates the average resistive force.
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A circular racetrack of radius 300 m is banked at an angle of 15°. If the coefficient of friction between the wheels of a race-car and the road is 0.2, what is the (a) optimum speed of the racecar to avoid wear and tear on its tyres, and (b) maximum permissible speed to avoid slipping ?
AnswerOn a banked road, the horizontal component of the normal force and the frictional force contribute to provide centripetal force to keep the car moving on a circular turn without slipping. At the optimum speed, the normal reaction's component is enough to provide the needed centripetal force, and the frictional force is not needed. The optimum speed $v_o$ is given by Eq. (4.22):
$
v_0=(R g \tan \theta)^{1 / 2}
$
Here $R=300 m , \theta=15^{\circ}, g=9.8 m s ^{-2}$; we have
$
v_0=28.1 m s ^{-1} .
$
The maximum permissible speed $v_{\max }$ is given by Eq. (4.21):
$
v_{\max }=\left(R g \frac{\mu_s+\tan \theta}{1-\mu_s \tan \theta}\right)^{1 / 2}=38.1 m s ^{-1}
$
View full question & answer→Question 672 Marks
A cyclist speeding at 18 km/h on a level road takes a sharp circular turn of radius 3 m without reducing the speed. The co-efficient of static friction
between the tyres and the road is 0.1. Will the cyclist slip while taking the turn?
AnswerOn an unbanked road, frictional force alone can provide the centripetal force needed to keep the cyclist moving on a circular turn without slipping. If the speed is too large, or if the turn is too sharp (i.e. of too small a radius) or both, the frictional force is not sufficient to provide the necessary centripetal force, and the cyclist slips. The condition for the cyclist not to slip is given by Eq. (4.18) :
$
v^2 \leq \mu_s R g
$
Now, $R=3 m , g=9.8 m s ^{-2}, \mu_s=0.1$. That is, $\mu_{ s } R g=2.94 m ^2 s ^{-2} . v=18 km / h =5 m s ^{-1}$;i.e., $v^2=25 m ^2 s ^{-2}$. The condition is not obeyed. The cyclist will slip while taking the circular turn.
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