NCERT Exemplar Solutions Class 11 Physics Laws of Motion — Multiple Choice Questions Q1–15

NCERT Exemplar Problems, Class 11 Physics — Laws of Motion, MCQ I and MCQ II. All 15 questions solved.

The book numbers these questions 5.1 to 5.15; here they are Questions 1 to 15. Questions 1 to 9 are MCQ I (exactly one option is correct) and Questions 10 to 15 are MCQ II (one or more options may be correct). Three ideas carry the whole page:

• Force is the rate of change of momentum. $\mathbf{F} = d\mathbf{p}/dt$, and the impulse $\int\mathbf{F}\,dt$ equals the change in momentum, whether the change is slow or sudden.
• Forces come in pairs. The third law makes the impulses in a collision equal and opposite, which is what conservation of momentum rests on.
• Friction is self-adjusting up to a limit. Static friction takes whatever value is needed, up to $\mu N$; when a problem asks “will it move?”, compare the force needed with that limit.

Key insight. In every two-body question (11, 12, 13) the deciding step is to ask what friction would have to do and whether it can. In Question 10 the deciding step is to differentiate carefully: the impulse is $m\,\Delta v$, and $\Delta v$ carries one factor of $4\pi$, not $4\pi^2$.

Choose the correct answer from the given options. (MCQ I: one correct. MCQ II: one or more correct.)

Question 1

A ball is travelling with uniform translatory motion. This means that

Solution. Translatory means every point of the body moves the same way, with no rotation; uniform means that common velocity does not change. Together they say all parts have the same constant velocity. A ball at rest (a) is not travelling. A circular path (b) has a changing direction, so the velocity is not uniform. A spinning ball (d) has points moving with different velocities, so it is not pure translation.

Question 2

A metre scale is moving with uniform velocity. This implies

Solution. Uniform velocity means no acceleration of the centre of mass, so by the second law the net force is zero. A scale moving with uniform velocity is translating without starting to rotate; a net torque about the centre of mass would give it an angular acceleration, so the torque must be zero as well.

Question 3

A cricket ball of mass $150$ g has an initial velocity $\mathbf{u} = (3\hat{\mathbf{i}} + 4\hat{\mathbf{j}})$ m s$^{-1}$ and a final velocity $\mathbf{v} = -(3\hat{\mathbf{i}} + 4\hat{\mathbf{j}})$ m s$^{-1}$ after being hit. The change in momentum (final momentum − initial momentum) is (in kg m s$^{-1}$)

Solution. Momentum is a vector, so subtract the vectors, not the speeds: $$\Delta\mathbf{p} = m(\mathbf{v} – \mathbf{u}) = 0.15\left[-(3\hat{\mathbf{i}} + 4\hat{\mathbf{j}}) – (3\hat{\mathbf{i}} + 4\hat{\mathbf{j}})\right] = -0.15(6\hat{\mathbf{i}} + 8\hat{\mathbf{j}}) = -(0.9\hat{\mathbf{i}} + 1.2\hat{\mathbf{j}}).$$ Option (a) is what one gets by comparing speeds, which are equal; option (b) forgets that the velocity reverses, so the change is twice $m\mathbf{u}$.

Question 4

In the previous problem (Question 3), the magnitude of the momentum transferred during the hit is

Solution. The momentum transferred is the magnitude of $\Delta\mathbf{p}$ from Question 3: $$|\Delta\mathbf{p}| = \sqrt{0.9^2 + 1.2^2} = \sqrt{2.25} = 1.5\ \text{kg m s}^{-1}.$$ Equivalently, the speed is $5$ m s$^{-1}$ before and after and the direction reverses, so $|\Delta p| = 2 \times 0.15 \times 5 = 1.5$.

Question 5

Conservation of momentum in a collision between particles can be understood from

Solution. During the collision, the second law says each particle’s change of momentum equals the impulse on it. The third law says the forces they exert on each other are equal and opposite at every instant, so the two impulses are equal and opposite. Adding, the total change of momentum is zero. Both laws are needed; energy conservation is a separate statement and is not even true in an inelastic collision.

Question 6

A hockey player is moving northward and suddenly turns westward with the same speed to avoid an opponent. The force that acts on the player is

Solution. The force is along the change in momentum. Take north as $+\hat{\mathbf{j}}$ and west as $-\hat{\mathbf{i}}$: $\Delta\mathbf{v} = v(-\hat{\mathbf{i}}) – v\hat{\mathbf{j}} = -v(\hat{\mathbf{i}} + \hat{\mathbf{j}})$, which points south-west. Muscles are internal forces and cannot change the momentum of the player as a whole; the external horizontal force that does is friction from the ice (or ground) on the skates, pushed into action by the player’s muscles.

Question 7

A body of mass $2$ kg travels according to the law $x(t) = pt + qt^2 + rt^3$ where $p = 3$ m s$^{-1}$, $q = 4$ m s$^{-2}$ and $r = 5$ m s$^{-3}$. The force acting on the body at $t = 2$ seconds is

Solution. Differentiate twice: $a = \dfrac{d^2x}{dt^2} = 2q + 6rt = 8 + 30t$. At $t = 2$ s, $a = 68$ m s$^{-2}$, so $$F = ma = 2 \times 68 = 136\ \text{N}.$$ Option (d) is the acceleration mistaken for the force.

Question 8

A body with mass $5$ kg is acted upon by a force $\mathbf{F} = (-3\hat{\mathbf{i}} + 4\hat{\mathbf{j}})$ N. If its initial velocity at $t = 0$ is $\mathbf{v} = (6\hat{\mathbf{i}} – 12\hat{\mathbf{j}})$ m s$^{-1}$, the time at which it will just have a velocity along the $y$-axis is

Solution. The acceleration is $\mathbf{a} = \mathbf{F}/m = (-0.6\hat{\mathbf{i}} + 0.8\hat{\mathbf{j}})$ m s$^{-2}$. The velocity points along the $y$-axis when its $x$-component vanishes: $$v_x = 6 – 0.6t = 0 \quad\Rightarrow\quad t = 10\ \text{s}.$$ At that moment $v_y = -12 + 8 = -4$ m s$^{-1}$, which is non-zero, so the velocity really is along the ($-y$) axis.

Question 9

A car of mass $m$ starts from rest and acquires a velocity along east $\mathbf{v} = v\hat{\mathbf{i}}$ $(v > 0)$ in two seconds. Assuming the car moves with uniform acceleration, the force exerted on the car is

Solution. The acceleration is $v/2$, so the net force is $mv/2$ eastward. The engine is part of the car: its forces are internal and cannot accelerate the car on their own. The engine turns the wheels, the tyres push backward on the road, and the road’s static friction pushes the car forward. That external friction is the force of $mv/2$.

Question 10

The motion of a particle of mass $m$ is given by $x = 0$ for $t < 0$ s, $x(t) = A\sin 4\pi t$ for $0 < t < (1/4)$ s $(A > 0)$, and $x = 0$ for $t > (1/4)$ s. Which of the following statements is true?

Solution. Differentiate in the middle interval: $$v = 4\pi A\cos 4\pi t, \qquad a = -16\pi^2 A\sin 4\pi t.$$

• (a) At $t = 1/8$ s, $\sin(\pi/2) = 1$, so $F = ma = -16\pi^2 A\,m$. True.
• (b) Just before $t = 0$ the particle is at rest; just after, $v = 4\pi A$. Just before $t = 1/4$ s, $v = 4\pi A\cos\pi = -4\pi A$; just after, it is at rest. So there are two sudden jumps in velocity, each of size $4\pi A$, and each is an impulse of magnitude $m\,\Delta v = 4\pi A\,m$. The option says $4\pi^2 A\,m$, which is wrong by a factor of $\pi$. False as printed.
• (c) The particle accelerates between $0$ and $1/4$ s, so a force acts. False.
• (d) The force $-16\pi^2 A m\sin 4\pi t$ varies with time, and there are impulses at the ends. True.
• (e) There are impulses at $t = 0$ and $t = 1/4$ s. False.

Printed key error. The key prints (a), (b), (d). The idea behind (b), that impulses act at $t = 0$ and $t = 1/4$ s, is right, but the magnitude it states is $4\pi^2 Am$ while the velocity jump is $4\pi A$, giving $4\pi A m$. As printed, (b) is false, so the answer is (a), (d).

Question 11

In Fig. 5.1, the coefficient of friction between the floor and the body B is $0.1$. The coefficient of friction between the bodies B and A is $0.2$. A force $\mathbf{F}$ is applied as shown on B. The mass of A is $m/2$ and of B is $m$. Which of the following statements are true?

The figure shows block A resting on top of block B, which lies on the floor; the horizontal force $F$ pulls B.

Solution. Two thresholds decide everything.

To start moving. The floor carries both blocks, normal force $1.5\,mg$, so the limiting floor friction is $0.1 \times 1.5\,mg = 0.15\,mg$. Below this, nothing moves.

To move together. A is dragged along only by friction from B, which can give A at most $0.2 \times \tfrac{m}{2}g$, an acceleration of at most $0.2g$. For the pair to share that acceleration, $$F – 0.15\,mg = 1.5\,m \times 0.2g \quad\Rightarrow\quad F_{\max} = 0.45\,mg.$$

Now test each force. $F = 0.1\,mg < 0.15\,mg$: at rest, so (d) is true. $F = 0.25\,mg$ lies between $0.15\,mg$ and $0.45\,mg$: they move together, so (a) is true. $F = 0.5\,mg > 0.45\,mg$: A slips on B, so (b) is true and (c) is false. And (e) is exactly the threshold found above.

Question 12

Mass $m_1$ moves on a slope making an angle $\theta$ with the horizontal and is attached to mass $m_2$ by a string passing over a frictionless pulley as shown in Fig. 5.2. The coefficient of friction between $m_1$ and the sloping surface is $\mu$. Which of the following statements are true?

The figure shows $m_1$ on the incline, the string running up the slope over a pulley at the top, and $m_2$ hanging vertically on the far side.

Solution. Friction always opposes the motion that would otherwise happen, and its largest value is $\mu m_1 g\cos\theta$.

• Up the plane. The pull $m_2g$ must beat both the component of weight $m_1g\sin\theta$ and the largest friction, which then acts down the slope: $m_2 > m_1(\sin\theta + \mu\cos\theta)$. So (b) is true. Option (a) ignores friction: a pull just above $m_1g\sin\theta$ can be held back by friction, so it does not guarantee motion. Option (c) is the opposite inequality and gives no upward motion.
• Down the plane. Now friction acts up the slope, so $m_1g\sin\theta$ must beat $m_2g + \mu m_1g\cos\theta$: $m_2 < m_1(\sin\theta – \mu\cos\theta)$. So (d) is true.

Question 13

In Fig. 5.3, a body A of mass $m$ slides on plane inclined at angle $\theta_1$ to the horizontal and $\mu_1$ is the coefficient of friction between A and the plane. A is connected by a light string passing over a frictionless pulley to another body B, also of mass $m$, sliding on a frictionless plane inclined at angle $\theta_2$ to the horizontal. Which of the following statements are true?

The figure shows a double incline with the pulley at the apex: A on the left face (angle $\theta_1$), B on the right face (angle $\theta_2$).

Solution. The masses are equal, so the string carries B’s pull $mg\sin\theta_2$ against A’s $mg\sin\theta_1$. For A to go up, B’s pull must also beat friction on A, which then acts down A’s slope: $$mg\sin\theta_2 \ge mg\sin\theta_1 + \mu_1 mg\cos\theta_1 \quad\Rightarrow\quad \mu_1 \le \frac{\sin\theta_2 – \sin\theta_1}{\cos\theta_1}.$$ A just starts to move up at equality, so (b) is true. Since $\mu_1 \ge 0$, this needs $\sin\theta_2 > \sin\theta_1$, that is $\theta_2 > \theta_1$, so (c) is true. (a) is then false: with a steep enough $\theta_2$, A does go up. (d) is false because, once moving, the net force is in general not zero, so the speed changes.

Question 14

Two billiard balls A and B, each of mass $50$ g and moving in opposite directions with speed of $5$ m s$^{-1}$ each, collide and rebound with the same speed. If the collision lasts for $10^{-3}$ s, which of the following statements are true?

Solution. Each ball reverses its velocity, so $$|\Delta p| = m(v + v) = 0.05 \times 10 = 0.5\ \text{N s},$$ and the average force is $0.5/10^{-3} = 500$ N. So (c) is true; (a) and (b) use $mv = 0.25$, forgetting that the velocity reverses. By the third law the force on A is equal and opposite to the force on B at every instant, so their impulses are too: (d) is true.

Question 15

A body of mass $10$ kg is acted upon by two perpendicular forces, $6$ N and $8$ N. The resultant acceleration of the body is

Solution. The resultant force is $\sqrt{6^2 + 8^2} = 10$ N, so $a = 10/10 = 1$ m s$^{-2}$, which rules out (b) and (d). Measured from the $6$ N force, the resultant makes an angle whose tangent is $8/6 = 4/3$; measured from the $8$ N force, the tangent is $6/8 = 3/4$. These describe the same direction, so (a) and (c) are both true.

Common mistakes

• Choosing (b) or (d) in Question 1. A circular path changes direction and a spinning ball has points with different velocities; neither is uniform translation.
• Subtracting speeds in Question 3. The speeds are equal but the velocity reverses, so $\Delta\mathbf{p} = -2m\mathbf{u}$, not zero.
• Calling the engine or the muscles the external force in Questions 6 and 9. Internal forces cannot change the momentum of the whole body; the road or ice supplies the force through friction.
• Ticking (b) in Question 10 because impulses do act. Check the number: the velocity jumps by $4\pi A$, so the impulse is $4\pi A m$, not $4\pi^2 A m$.
• Forgetting the floor friction in Question 11. The threshold for moving together is $0.45\,mg$, not $0.3\,mg$, because $0.15\,mg$ goes on floor friction first.
• Leaving friction out in Question 12(a). $m_2 > m_1\sin\theta$ is not enough when friction can hold the block.
• Using $mv$ instead of $2mv$ for the impulse in Question 14. Rebounding with the same speed doubles the change in momentum.
• Rejecting (c) in Question 15 as a different direction. $\tan^{-1}(4/3)$ from one force and $\tan^{-1}(3/4)$ from the other are the same line.

Practise next

• A $0.2$ kg ball moving at $10$ m s$^{-1}$ strikes a wall normally and rebounds at $8$ m s$^{-1}$. The contact lasts $0.02$ s. Find the average force on the ball.

• Block A ($2$ kg) rests on block B ($4$ kg), which lies on a frictionless floor. The coefficient of friction between A and B is $0.3$. A horizontal force $F$ pulls B. Find the largest $F$ for which the blocks move together ($g = 10$ m s$^{-2}$).

• Exemplar Laws of Motion: Very Short and Short Answer Questions — the same laws applied to lifts, recoil, impulse graphs and strings.
• Exemplar Laws of Motion: Long Answer Questions — inclines, circular tracks and equilibrium of forces at a point.
• NCERT textbook Chapter 4 — Laws of Motion for the theory of the three laws, momentum, friction and circular motion.