NCERT Exemplar Solutions Class 11 Physics Laws of Motion — Very Short and Short Answer Questions Q16–33

NCERT Exemplar Problems, Class 11 Physics — Laws of Motion, Very Short Answer Questions 16 to 25 and Short Answer Questions 26 to 33. All 18 questions solved. The book numbers them 5.16 to 5.33.

Three ideas carry this page. Momentum is conserved when no external force acts, which settles the bicycle, the gun and every recoil question. Force is how quickly momentum changes, so stretching the stopping time (straw, mud, a seat belt) lowers the force for the same change. In an accelerating frame, the apparent weight changes: a lift accelerating upwards adds $ma$ to the normal force, one falling with acceleration $a$ removes it.

Key insight. Before writing any equation, decide which body you are applying the law to and list only the external forces on it. Questions 26 and 28 are just the second law applied to one block at a time.

Question 16

A girl riding a bicycle along a straight road with a speed of $5$ m s$^{-1}$ throws a stone of mass $0.5$ kg which has a speed of $15$ m s$^{-1}$ with respect to the ground along her direction of motion. The mass of the girl and bicycle is $50$ kg. Does the speed of the bicycle change after the stone is thrown? What is the change in speed, if so?

Solution. No horizontal external force acts during the throw (friction is negligible over that short time), so horizontal momentum is conserved. Before the throw, the girl, bicycle and stone move together: $$p_i = (50 + 0.5) \times 5 = 252.5\ \text{kg m s}^{-1}.$$ After it, with $v$ the new speed of the bicycle, $$p_f = 50v + 0.5 \times 15 = 50v + 7.5.$$ Equating, $50v = 245$, so $v = 4.9$ m s$^{-1}$. Throwing the stone forward pushes the girl back, so the bicycle slows down.

Question 17

A person of mass $50$ kg stands on a weighing scale on a lift. If the lift is descending with a downward acceleration of $9$ m s$^{-2}$, what would be the reading of the weighing scale? $(g = 10$ m s$^{-2})$

Solution. Let $R$ be the push of the scale on the person. Taking downwards as positive, $$mg – R = ma \quad\Rightarrow\quad R = m(g – a) = 50(10 – 9) = 50\ \text{N}.$$ A scale calibrated in kilograms divides by $g$, so it shows $50/10 = 5$ kg.

Question 18

The position time graph of a body of mass $2$ kg is given in Fig. 5.4. What is the impulse on the body at $t = 0$ s and $t = 4$ s?

Solution. The graph is a straight line from $(0, 0)$ to $(4\ \text{s}, 3\ \text{m})$, so between $t = 0$ and $t = 4$ s the body moves with constant velocity $$v = \frac{3}{4} = 0.75\ \text{m s}^{-1}.$$ After $t = 4$ s the graph is flat, so the body is at rest.

At $t = 4$ s. The velocity drops suddenly from $0.75$ m s$^{-1}$ to zero, so the impulse is $$J = m(0 – 0.75) = 2 \times (-0.75) = -1.5\ \text{kg m s}^{-1},$$ that is, $1.5$ kg m s$^{-1}$ opposite to the motion.

At $t = 0$. The graph starts at $t = 0$ and shows nothing earlier. The key reads it as the body already moving at $0.75$ m s$^{-1}$ when the record starts, so there is no change of velocity at $t = 0$ and the impulse is zero. (If instead the body is taken to be at rest before $t = 0$, as in the similar textbook exercise whose graph shows a flat line for $t < 0$, the impulse at $t = 0$ would be $+1.5$ kg m s$^{-1}$.)

Question 19

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?

Solution. By inertia (the first law), the driver’s body keeps moving forward at the car’s speed unless an external force stops it. When the brakes stop the car, the only force acting backward on him is the friction from the seat, which is far too small to stop his upper body in the same short time. So he keeps going forward relative to the car and hits the steering wheel. A seat belt supplies the large backward force that friction cannot.

Question 20

The velocity of a body of mass $2$ kg as a function of $t$ is given by $\mathbf{v}(t) = 2t\,\hat{\mathbf{i}} + t^2\,\hat{\mathbf{j}}$. Find the momentum and the force acting on it, at time $t = 2$ s.

Solution. At $t = 2$ s, $\mathbf{v} = 4\hat{\mathbf{i}} + 4\hat{\mathbf{j}}$, so $$\mathbf{p} = m\mathbf{v} = 8\hat{\mathbf{i}} + 8\hat{\mathbf{j}}\ \text{kg m s}^{-1}.$$ The acceleration is $\mathbf{a} = d\mathbf{v}/dt = 2\hat{\mathbf{i}} + 2t\,\hat{\mathbf{j}}$, which at $t = 2$ s is $2\hat{\mathbf{i}} + 4\hat{\mathbf{j}}$. Hence $$\mathbf{F} = m\mathbf{a} = 4\hat{\mathbf{i}} + 8\hat{\mathbf{j}}\ \text{N}.$$

Question 21

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 $F$.

Solution. While the block stays at rest, static friction exactly balances the pull, so $f = F$: a straight line at $45^\circ$ through the origin. This holds up to the limiting value $f_s = \mu_s N$. Once $F$ exceeds that, the block slides and friction becomes kinetic, $f_k = \mu_k N$, which is a little smaller and does not change however hard you pull. So the graph rises along $f = F$, drops slightly at the start of sliding, and then runs flat.

The key draws the same graph without the small drop, which is the idealisation $\mu_s = \mu_k$.

Question 22

Why are porcelain objects wrapped in paper or straw before packing for transportation?

Solution. A truck may have to stop or jolt suddenly, and the porcelain must then lose its momentum. The force needed is $F = \Delta p/\Delta t$. Packed tightly against a hard surface, the object stops almost instantly, so $\Delta t$ is tiny and the force is large enough to break it. Straw or paper is soft and gives way, so the object moves a little further and takes longer to stop; the same change of momentum then needs a much smaller force.

Question 23

Why does a child feel more pain when she falls down on a hard cement floor, than when she falls on the soft muddy ground in the garden?

Solution. In both falls the child arrives with the same momentum and is brought to rest, so the change in momentum is the same. Cement does not yield, so she stops in a very short time and the force $\Delta p/\Delta t$ is large. Mud gives way: she sinks in a little and comes to rest over a longer time, so the force on her body is smaller and she feels less pain.

Question 24

A woman throws an object of mass $500$ g with a speed of $25$ m s$^{-1}$.

(a) What is the impulse imparted to the object?

(b) If the object hits a wall and rebounds with half the original speed, what is the change in momentum of the object?

Solution. (a) The object starts from rest, so the impulse is its final momentum: $$J = mv = 0.5 \times 25 = 12.5\ \text{N s}.$$

(b) Take the initial direction as positive. The object arrives with $+25$ m s$^{-1}$ and leaves with $-12.5$ m s$^{-1}$, so $$\Delta p = 0.5(-12.5 – 25) = -18.75\ \text{kg m s}^{-1}.$$ The magnitude is $18.75$ kg m s$^{-1}$, directed away from the wall.

Question 25

Why are mountain roads generally made winding upwards rather than going straight up?

Solution. On a road sloping at angle $\theta$, the friction available to grip the tyres is $f = \mu N = \mu mg\cos\theta$, while the component of weight pulling the vehicle down the slope is $mg\sin\theta$. A road straight up a mountain would be steep: $\cos\theta$ would be smaller, so less friction is available, and $\sin\theta$ would be larger, so more is needed. Vehicles would skid. Winding the road round the mountain covers the same height over a longer distance, keeping $\theta$ small, so friction is larger and the chance of skidding is less. A gentler slope also needs less engine force to climb.

Question 26

A mass of $2$ kg is suspended with thread AB (Fig. 5.5). Thread CD of the same type is attached to the other end of $2$ kg mass. Lower thread is pulled gradually, harder and harder in the downward direction so as to apply force on AB. Which of the threads will break and why?

The figure shows thread AB hanging from the ceiling, the $2$ kg mass hanging from B, and thread CD hanging below the mass from C to D.

Solution. With a slow, steady pull $F$ on CD, the mass stays in equilibrium at every stage. The tension in CD is $F$. The tension in AB must hold up both the weight and the pull: $$T_{AB} = F + mg = F + 20\ \text{N}.$$ AB therefore always carries $20$ N more than CD. As $F$ increases, AB reaches the breaking tension first, so the upper thread breaks.

Question 27

In the above given problem if the lower thread is pulled with a jerk, what happens?

Solution. A jerk applies a large force for a very short time. The $2$ kg mass has inertia: in that brief time it barely moves, so the upper thread is hardly stretched further and the extra pull does not reach AB. The whole of the large force acts on CD, which reaches its breaking tension first. So CD breaks.

Question 28

Two masses of $5$ kg and $3$ kg are suspended with help of massless inextensible strings as shown in Fig. 5.6. Calculate $T_1$ and $T_2$ when whole system is going upwards with acceleration $= 2$ m s$^{-2}$ (use $g = 9.8$ m s$^{-2}$).

The figure shows a string with tension $T_1$ from the ceiling to the $5$ kg mass, and a second string with tension $T_2$ from the $5$ kg mass down to the $3$ kg mass.

Solution. Apply the second law to one body at a time, upwards positive.

The $3$ kg mass. Only $T_2$ and its weight act: $$T_2 – 3g = 3a \quad\Rightarrow\quad T_2 = 3(9.8 + 2) = 35.4\ \text{N}.$$

Both masses together. $T_1$ and the total weight act: $$T_1 – 8g = 8a \quad\Rightarrow\quad T_1 = 8(9.8 + 2) = 94.4\ \text{N}.$$

Question 29

Block A of weight $100$ N rests on a frictionless inclined plane of slope angle $30^\circ$ (Fig. 5.7). A flexible cord attached to A passes over a frictionless pulley and is connected to block B of weight $W$. Find the weight $W$ for which the system is in equilibrium.

The figure shows A on the incline with the cord running up the slope over a pulley at the top and B hanging vertically.

Solution. In equilibrium the cord tension equals $W$ (B hangs at rest). Along the frictionless incline, the only force pulling A down the slope is the component of its weight, $100\sin 30^\circ$. Balancing, $$W = 100\sin 30^\circ = 50\ \text{N}.$$

Question 30

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.

Solution. The finger pushes horizontally with force $F$, and the wall pushes back with normal force $N = F$. The block’s weight $Mg$ must be held by friction, which acts upwards and can be at most $\mu N = \mu F$. For the block not to slip, $$\mu F \ge Mg \quad\Rightarrow\quad F_{\min} = \frac{Mg}{\mu}.$$

Question 31

A $100$ kg gun fires a ball of $1$ kg horizontally from a cliff of height $500$ m. It falls on the ground at a distance of $400$ m from the bottom of the cliff. Find the recoil velocity of the gun. (acceleration due to gravity $= 10$ m s$^{-2}$)

Solution. First find the ball’s launch speed from its flight. Falling $500$ m from rest vertically takes $$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{1000}{10}} = 10\ \text{s},$$ and in that time it travels $400$ m horizontally, so $u = 400/10 = 40$ m s$^{-1}$. During the firing, horizontal momentum is conserved and is zero to start with: $$100V = 1 \times 40 \quad\Rightarrow\quad V = 0.4\ \text{m s}^{-1},$$ opposite to the direction of the ball.

Question 32

Figure 5.8 shows $(x, t)$, $(y, t)$ diagram of a particle moving in 2-dimensions. If the particle has a mass of $500$ g, find the force (direction and magnitude) acting on the particle.

Fig. 5.8(a) shows $x$ against $t$ as a straight line through the origin. Fig. 5.8(b) shows $y$ against $t$ as a parabola through the origin, passing through $y = 4$ m at $t = 2$ s.

Solution. A straight $x$–$t$ line means constant $v_x$, so $a_x = 0$ (the key reads it as $x = t$; the slope does not matter for the force). The $y$–$t$ curve is a parabola from rest at the origin through $(2\ \text{s}, 4\ \text{m})$, so $y = t^2$ and $$a_y = \frac{d^2y}{dt^2} = 2\ \text{m s}^{-2}.$$ The force is therefore entirely along $y$: $$F = ma_y = 0.5 \times 2 = 1\ \text{N}.$$

Question 33

A person in an elevator accelerating upwards with an acceleration of $2$ m s$^{-2}$, tosses a coin vertically upwards with a speed of $20$ m s$^{-1}$. After how much time will the coin fall back into his hand? $(g = 10$ m s$^{-2})$

Solution. Work in the elevator’s frame. Relative to the person, the coin starts at $20$ m s$^{-1}$ upwards and accelerates downwards at $g + a = 12$ m s$^{-2}$ (gravity, plus the elevator accelerating up to meet it). It returns to the hand when its relative displacement is zero: $$0 = 20t – \tfrac12(12)t^2 \quad\Rightarrow\quad t = \frac{2 \times 20}{12} = \frac{10}{3} \approx 3.33\ \text{s}.$$

Common mistakes

• Leaving the stone’s mass out of the starting momentum in Question 16. Before the throw it is $50.5$ kg moving at $5$ m s$^{-1}$, not $50$ kg.
• Using $g + a$ for a descending lift in Question 17. The lift is accelerating downwards, so the normal force is $m(g – a)$.
• Taking $mv$ instead of the change in momentum in Question 24(b). The rebound reverses the velocity, so the speeds add.
• Treating the gradual pull and the jerk alike in Questions 26 and 27. Slowly, the weight adds to AB’s load; suddenly, inertia shields AB.
• Using $T_1 = 5(g + a)$ in Question 28. $T_1$ supports both masses; it is $8(g + a)$.
• Taking $F = \mu Mg$ in Question 30. The finger’s push is the normal force; friction $\mu F$ must equal $Mg$, so $F = Mg/\mu$.
• Using $g$ alone for the coin in Question 33. Relative to the person, the coin’s acceleration is $g + a$.

Practise next

• A $60$ kg person stands on a weighing scale in a lift that is accelerating upwards at $2$ m s$^{-2}$. What does the scale read, in kilograms? ($g = 10$ m s$^{-2}$)

• A $4$ kg block hangs from the ceiling of a lift by one string, and a $2$ kg block hangs from the $4$ kg block by a second string. The lift accelerates upwards at $3$ m s$^{-2}$. Find the tensions in the upper and lower strings. ($g = 10$ m s$^{-2}$)

• Exemplar Laws of Motion: Multiple Choice Questions — the same laws in objective form, including friction between stacked blocks.
• Exemplar Laws of Motion: Long Answer Questions — inclines, circular tracks and the helicopter problem.
• NCERT textbook Chapter 4 — Laws of Motion for the theory of the three laws, momentum, friction and circular motion.