NCERT Solutions Class 11 Physics Chapter 4 — Laws of Motion

NCERT Class 11 Physics — Laws of Motion, Chapter 4 Exercises. All 23 questions solved.

The exercises at the end of Chapter 4 fall into four groups. Questions 4.1 to 4.4, 4.12 and 4.22 ask for the net force on a body, and what a body does when a force stops acting: the first law. Questions 4.5 to 4.11, 4.13 to 4.16 and 4.21 put numbers into the second law, for single bodies, a lift, connected bodies and a stone on a string. Questions 4.17 to 4.20 are about momentum: impulse, recoil and conservation of momentum. Question 4.23 asks for four everyday effects to be explained. The whole chapter runs on three results:

• The second law — $\vec F_{\text{net}} = m\vec a$, or more generally $\vec F = \dfrac{d\vec p}{dt}$. The net force fixes the acceleration, not the velocity.
• Impulse and momentum — impulse $= \vec F\,\Delta t = \Delta\vec p = m\vec v – m\vec u$. With no external force on a system, its total momentum is conserved.
• Circular motion — a body moving at speed $v$ in a circle of radius $r$ needs a net force $mv^2/r$ towards the centre, supplied by some real force such as the tension in a string.

The exercises say to take $g = 10\ \text{m s}^{-2}$, and that value is used throughout.

Key insight. The net force on a body depends only on its acceleration at that instant — never on its velocity, and never on how it got there. A car at constant velocity has no net force on it, however fast it goes (4.1). A pebble at the top of its flight, momentarily at rest, still has its full weight acting (4.2). A stone released from an accelerating train keeps none of the train’s push (4.3). So in every question, pick one body, draw the forces on it, add them, and set the sum equal to $m\vec a$.

Question 4.1

Give the magnitude and direction of the net force acting on (a) a drop of rain falling down with a constant speed, (b) a cork of mass $10\ \text{g}$ floating on water, (c) a kite skillfully held stationary in the sky, (d) a car moving with a constant velocity of $30\ \text{km/h}$ on a rough road, (e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields.

Solution. The first law says a body with no net force on it keeps moving with constant velocity. The second law turns this round: if the velocity is constant, the acceleration is zero, so the net force must be zero. Every part is a case of constant velocity, including zero velocity.

In (a) the raindrop falls at constant speed, so its weight is exactly balanced by the upward drag of the air. In (b) the cork is at rest; its weight, $mg = 0.010\times10 = 0.1\ \text{N}$, is balanced by the upward buoyant force of the water. In (c) the kite is stationary, so its weight, the pull of the string and the force of the wind on it add to zero. In (d) the car’s velocity is constant in size and direction, so the forward force of the road on the driving wheels exactly balances friction and air resistance. The roughness of the road changes the size of those forces, not their sum. In (e) no force acts on the electron at all, since there is no matter and no field nearby to exert one. It moves with constant velocity however fast it is going.

Question 4.2

A pebble of mass $0.05\ \text{kg}$ is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble, (a) during its upward motion, (b) during its downward motion, (c) at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of $45^\circ$ with the horizontal direction? Ignore air resistance.

Solution. Once the pebble has left the hand, the only force on it is gravity; nothing pushes it upwards any more. So the net force throughout is its weight,

$$F = mg = 0.05\times10 = 0.5\ \text{N}$$

vertically downwards, in all three stages. The pebble’s velocity is upwards in (a), downwards in (b) and zero in (c), but the force does not depend on the velocity, only on what acts on the pebble, and that does not change.

Thrown at $45^\circ$. The force is the same: the weight, $0.5\ \text{N}$ vertically downwards, at every point of the flight. The one difference is at the highest point. The pebble is not at rest there, because it keeps the horizontal component of its velocity throughout the flight, as there is no horizontal force to change it.

Question 4.3

Give the magnitude and direction of the net force acting on a stone of mass $0.1\ \text{kg}$, (a) just after it is dropped from the window of a stationary train, (b) just after it is dropped from the window of a train running at a constant velocity of $36\ \text{km/h}$, (c) just after it is dropped from the window of a train accelerating with $1\ \text{m s}^{-2}$, (d) lying on the floor of a train which is accelerating with $1\ \text{m s}^{-2}$, the stone being at rest relative to the train. Neglect air resistance throughout.

Solution. In (a) to (c) the words that matter are just after. Once the stone leaves the window it is not touching the train, so the train can exert no force on it. With air resistance neglected, the only force is its weight, $mg = 0.1\times10 = 1\ \text{N}$, downwards. What the train was doing is irrelevant: the force at an instant depends on the situation at that instant, not on the stone’s history. In (b) the stone does carry the train’s horizontal velocity of $10\ \text{m s}^{-1}$ with it, and in (c) whatever velocity the train had at the moment of release, but keeping a velocity needs no force.

In (d) the stone is in contact with the floor and moves with the train, so it shares the train’s acceleration of $1\ \text{m s}^{-2}$. Its weight is balanced by the normal force of the floor. What is left is the horizontal force of the floor on it, friction, and that is the net force:

$$F = ma = 0.1\times1 = 0.1\ \text{N}$$

Question 4.4

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].

Solution. The particle’s weight and the normal force of the table are vertical and cancel. The only horizontal force is the tension, which pulls the particle towards the peg at the centre of the circle, so the net force towards the centre is simply $T$. It is this tension that provides the centripetal force, $T = mv^2/l$. The centripetal force is not an extra force to be added to or taken away from $T$, so options (ii) and (iii) count the same force twice. Option (iv) would leave the particle nothing to turn it, and it would move in a straight line.

Question 4.5

A constant retarding force of $50\ \text{N}$ is applied to a body of mass $20\ \text{kg}$ moving initially with a speed of $15\ \text{m s}^{-1}$. How long does the body take to stop?

Solution. The force is constant, so the acceleration is constant and the equations of uniformly accelerated motion apply. The retarding force points against the motion, so the acceleration is negative:

$$a = \frac{F}{m} = \frac{-50}{20} = -2.5\ \text{m s}^{-2}$$

The body stops when $v = 0$, so from $v = u + at$:

$$0 = 15 – 2.5\,t \quad\Rightarrow\quad t = 6.0\ \text{s}$$

Question 4.6

A constant force acting on a body of mass $3.0\ \text{kg}$ changes its speed from $2.0\ \text{m s}^{-1}$ to $3.5\ \text{m s}^{-1}$ in $25\ \text{s}$. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

Solution. The direction of motion does not change, so the force acts along the line of motion and the problem is one-dimensional. The speed increases, so the force points forwards. The acceleration is the change in velocity divided by the time taken:

$$a = \frac{3.5 – 2.0}{25} = 0.06\ \text{m s}^{-2}, \qquad F = ma = 3.0\times0.06 = 0.18\ \text{N}$$

Question 4.7

A body of mass $5\ \text{kg}$ is acted upon by two perpendicular forces $8\ \text{N}$ and $6\ \text{N}$. Give the magnitude and direction of the acceleration of the body.

Solution. Forces are vectors, so two perpendicular forces combine by Pythagoras, not by simple addition:

$$F = \sqrt{8^2 + 6^2} = 10\ \text{N}$$

The resultant makes an angle $\theta$ with the $8\ \text{N}$ force, where $\tan\theta = 6/8 = 3/4$, so $\theta \approx 37^\circ$. The acceleration points along the net force and has magnitude

$$a = \frac{F}{m} = \frac{10}{5} = 2\ \text{m s}^{-2}$$

Question 4.8

The driver of a three-wheeler moving with a speed of $36\ \text{km/h}$ sees a child standing in the middle of the road and brings his vehicle to rest in $4.0\ \text{s}$ just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is $400\ \text{kg}$ and the mass of the driver is $65\ \text{kg}$.

Solution. The driver slows down with the vehicle, so the force must stop both, a total mass of $400 + 65 = 465\ \text{kg}$. The initial speed is $36\ \text{km/h} = 36\times\dfrac{1000}{3600} = 10\ \text{m s}^{-1}$, and it falls to zero in $4.0\ \text{s}$:

$$a = \frac{0 – 10}{4.0} = -2.5\ \text{m s}^{-2}, \qquad |F| = 465\times2.5 = 1162.5\ \text{N} \approx 1.2\times10^3\ \text{N}$$

Question 4.9

A rocket with a lift-off mass $20{,}000\ \text{kg}$ is blasted upwards with an initial acceleration of $5.0\ \text{m s}^{-2}$. Calculate the initial thrust (force) of the blast.

Solution. Two forces act on the rocket at lift-off: the thrust $F$ upwards and its weight $mg$ downwards. The net upward force produces the upward acceleration:

$$F – mg = ma \quad\Rightarrow\quad F = m(g + a) = 20000\times(10 + 5.0) = 3.0\times10^5\ \text{N}$$

The thrust has first to support the weight and then to provide the acceleration on top of that, which is why $g$ and $a$ add.

Question 4.10

A body of mass $0.40\ \text{kg}$ moving initially with a constant speed of $10\ \text{m s}^{-1}$ to the north is subject to a constant force of $8.0\ \text{N}$ directed towards the south for $30\ \text{s}$. Take the instant the force is applied to be $t = 0$, the position of the body at that time to be $x = 0$, and predict its position at $t = -5\ \text{s}$, $25\ \text{s}$, $100\ \text{s}$.

Solution. Take north as positive. The body starts with $u = +10\ \text{m s}^{-1}$, and while the force acts, for $0 \le t \le 30\ \text{s}$, its acceleration is

$$a = \frac{-8.0}{0.40} = -20\ \text{m s}^{-2}$$

The force acts for those $30\ \text{s}$ only, so each time asked for has to be placed in the right stage of the motion.

At $t = -5\ \text{s}$. Before the force was applied, the body was moving at a constant $10\ \text{m s}^{-1}$ north, so $5\ \text{s}$ earlier it was $50\ \text{m}$ south of $x = 0$:

$$x = ut = 10\times(-5) = -50\ \text{m}$$

At $t = 25\ \text{s}$. The force is still acting, so the equation of uniform acceleration applies:

$$x = ut + \tfrac12at^2 = 10\times25 – \tfrac12\times20\times25^2 = 250 – 6250 = -6000\ \text{m}$$

The body has turned round and is now $6\ \text{km}$ south of its starting point.

At $t = 100\ \text{s}$. The same equation cannot be used for the whole $100\ \text{s}$, because the force stops at $30\ \text{s}$. Up to $30\ \text{s}$:

$$x_1 = 10\times30 – \tfrac12\times20\times30^2 = -8700\ \text{m}, \qquad v = 10 – 20\times30 = -590\ \text{m s}^{-1}$$

After that no force acts, so by the first law the body keeps moving at $590\ \text{m s}^{-1}$ southwards for the remaining $70\ \text{s}$:

$$x_2 = -590\times70 = -41300\ \text{m}, \qquad x = x_1 + x_2 = -50000\ \text{m}$$

Question 4.11

A truck starts from rest and accelerates uniformly at $2.0\ \text{m s}^{-2}$. At $t = 10\ \text{s}$, a stone is dropped by a person standing on the top of the truck ($6\ \text{m}$ high from the ground). What are the (a) velocity, and (b) acceleration of the stone at $t = 11\ \text{s}$? (Neglect air resistance.)

Question 4.11 (a)

Solution. At $t = 10\ \text{s}$ the truck, and the stone with it, is moving at $v = 0 + 2.0\times10 = 20\ \text{m s}^{-1}$. When the stone is dropped it leaves with this horizontal velocity. From then on the only force on it is gravity, which is vertical. So by the first law its horizontal velocity stays $20\ \text{m s}^{-1}$; the truck’s acceleration no longer affects it.

First check that the stone is still in the air at $t = 11\ \text{s}$: in $1\ \text{s}$ it falls $\tfrac12\times10\times1^2 = 5\ \text{m}$, less than the $6\ \text{m}$ height. Its vertical velocity after $1\ \text{s}$ of falling is $10\times1 = 10\ \text{m s}^{-1}$ downwards. Combining the two perpendicular components:

$$v = \sqrt{20^2 + 10^2} = \sqrt{500} \approx 22.4\ \text{m s}^{-1}, \qquad \tan\theta = \frac{10}{20} = \frac12$$

where $\theta$ is measured below the horizontal.

Question 4.11 (b)

Solution. Once released, the stone has only its weight acting on it, so its acceleration is $g$, whatever its velocity.

Question 4.12

A bob of mass $0.1\ \text{kg}$ hung from the ceiling of a room by a string $2\ \text{m}$ long is set into oscillation. The speed of the bob at its mean position is $1\ \text{m s}^{-1}$. 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.

Solution. Once the string is cut, the only force on the bob is gravity, so it moves like any projectile. Its path is decided by its velocity at the instant of cutting. The mass and the length of the string do not come into it.

Question 4.12 (a)

Solution. At an extreme position the bob is momentarily at rest. With zero velocity and only its weight acting, it falls straight down.

Question 4.12 (b)

Solution. At the mean position the bob is moving horizontally at $1\ \text{m s}^{-1}$. A horizontal velocity combined with a constant vertical force gives a parabola, like a ball rolling off the edge of a table.

Question 4.13

A man of mass $70\ \text{kg}$ stands on a weighing scale in a lift which is moving (a) upwards with a uniform speed of $10\ \text{m s}^{-1}$, (b) downwards with a uniform acceleration of $5\ \text{m s}^{-2}$, (c) upwards with a uniform acceleration of $5\ \text{m s}^{-2}$. What would be the readings on the scale in each case? (d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?

Solution. A weighing scale measures the force with which the man presses down on it. By the third law this equals the normal force $N$ that the scale exerts upwards on him, so it is $N$ that has to be found. Two forces act on the man: $N$ upwards and his weight $mg = 70\times10 = 700\ \text{N}$ downwards. He moves with the lift, so he has its acceleration. Taking upwards as positive, the second law gives

$$N – mg = ma \quad\Rightarrow\quad N = m(g + a)$$

with $a$ negative for a downward acceleration. The scale is marked in kilograms, so it shows $N/g$.

Question 4.13 (a)

Solution. A uniform speed means $a = 0$, whichever way the lift is moving, so $N = mg = 700\ \text{N}$. Riding at a steady speed feels exactly like standing still.

Question 4.13 (b)

Solution. The acceleration is downwards, so $a = -5\ \text{m s}^{-2}$:

$$N = 70\times(10 – 5) = 350\ \text{N}$$

The floor gives way beneath him, so it needs to push up less than his weight.

Question 4.13 (c)

Solution. Now $a = +5\ \text{m s}^{-2}$, and the floor has to push harder than his weight to speed him up:

$$N = 70\times(10 + 5) = 1050\ \text{N}$$

Question 4.13 (d)

Solution. In free fall the lift, the scale and the man all accelerate downwards at $g$, so $a = -10\ \text{m s}^{-2}$ and $N = 70\times(10 – 10) = 0$. The man and the scale fall together without pressing on each other: he is weightless.

Question 4.14

Figure 4.16 shows the position-time graph of a particle of mass $4\ \text{kg}$. What is the (a) force on the particle for $t < 0$, $t > 4\ \text{s}$, $0 < t < 4\ \text{s}$? (b) impulse at $t = 0$ and $t = 4\ \text{s}$? (Consider one-dimensional motion only).

Question 4.14 (a)

Solution. Read the motion off the graph. For $t < 0$ the graph is flat at $x = 0$: the particle is at rest. For $0 < t < 4\ \text{s}$ it is a straight line rising from $0$ to $3\ \text{m}$, so the particle moves with a constant velocity of $3/4 = 0.75\ \text{m s}^{-1}$. For $t > 4\ \text{s}$ it is flat again at $x = 3\ \text{m}$: at rest once more. In each interval the velocity is constant, so the acceleration is zero, and so is the force.

Question 4.14 (b)

Solution. At $t = 0$ and at $t = 4\ \text{s}$ the slope of the graph, the velocity, changes suddenly. That takes a very large force acting for a very short time, which is an impulse, and an impulse equals the change in momentum it produces:

$$J_{t=0} = m(0.75 – 0) = 4\times0.75 = 3\ \text{kg m s}^{-1}, \qquad J_{t=4\ \text{s}} = m(0 – 0.75) = -3\ \text{kg m s}^{-1}$$

The first starts the particle moving; the second, in the opposite direction, stops it.

Question 4.15

Two bodies of masses $10\ \text{kg}$ and $20\ \text{kg}$ respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force $F = 600\ \text{N}$ is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case?

Solution. The surface is smooth, so the only horizontal forces are $F$ and the tension. Take the two bodies and the string together as one system. The tension is then an internal force, pulling the two bodies towards each other equally, and the whole $30\ \text{kg}$ accelerates under $F$ alone:

$$a = \frac{F}{m_A + m_B} = \frac{600}{10 + 20} = 20\ \text{m s}^{-2}$$

The tension is the only horizontal force on the trailing body, the one not being pulled, so it must give that body this acceleration. Its value therefore depends on which body is pulled.

In case (i) the force is applied to A, and the string drags B ($20\ \text{kg}$) behind it: $T = 20\times20 = 400\ \text{N}$. As a check, on A the net force is $600 – 400 = 200\ \text{N} = 10\times20$, as it should be.

In case (ii) the force is applied to B, and the string drags A ($10\ \text{kg}$): $T = 10\times20 = 200\ \text{N}$.

Question 4.16

Two masses $8\ \text{kg}$ and $12\ \text{kg}$ are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.

Solution. The heavier $12\ \text{kg}$ mass goes down and the $8\ \text{kg}$ mass goes up. The string is inextensible, so they move with the same acceleration $a$; it is light and the pulley frictionless, so the tension $T$ is the same on both sides. Write the second law for each mass, taking each one’s own direction of motion as positive:

$$T – 8\times10 = 8a, \qquad 12\times10 – T = 12a$$

Adding the two equations removes $T$: $40 = 20a$, so $a = 2\ \text{m s}^{-2}$. Then $T = 80 + 8\times2 = 96\ \text{N}$.

The tension lies between the two weights, $80\ \text{N}$ and $120\ \text{N}$, as it must: it has to exceed the lighter weight to lift that mass, and fall short of the heavier weight to let that one drop.

Question 4.17

A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.

Solution. The forces that break the nucleus up are internal, and no external force acts, so the total momentum is conserved. The nucleus was at rest, so the total momentum before the break-up is zero, and it must still be zero afterwards:

$$m_1\vec v_1 + m_2\vec v_2 = 0 \quad\Rightarrow\quad \vec v_1 = -\frac{m_2}{m_1}\,\vec v_2$$

Two momentum vectors can add to zero only if they are equal in size and opposite in direction. So the two products move along the same line in opposite directions, with speeds in the inverse ratio of their masses; the lighter one moves faster.

Question 4.18

Two billiard balls each of mass $0.05\ \text{kg}$ moving in opposite directions with speed $6\ \text{m s}^{-1}$ collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?

Solution. Follow one ball, moving to the right at $6\ \text{m s}^{-1}$ before the collision and to the left at $6\ \text{m s}^{-1}$ after it. Taking right as positive, the impulse on it is its change in momentum:

$$J = m(v – u) = 0.05\times(-6 – 6) = -0.6\ \text{kg m s}^{-1}$$

The speed is the same before and after, but the velocity has reversed, so the momentum changes by twice $mv$, not by zero. The other ball receives an equal and opposite impulse, as the third law requires.

Question 4.19

A shell of mass $0.020\ \text{kg}$ is fired by a gun of mass $100\ \text{kg}$. If the muzzle speed of the shell is $80\ \text{m s}^{-1}$, what is the recoil speed of the gun?

Solution. Before firing, the gun and the shell are at rest, so their total momentum is zero. The forces of the explosion are internal to the gun and shell, so the total momentum stays zero, and the gun must move backwards with a momentum equal in size to the shell’s:

$$100\,V = 0.020\times80 \quad\Rightarrow\quad V = 0.016\ \text{m s}^{-1}$$

Question 4.20

A batsman deflects a ball by an angle of $45^\circ$ without changing its initial speed which is equal to $54\ \text{km/h}$. What is the impulse imparted to the ball? (Mass of the ball is $0.15\ \text{kg}$.)

Solution. The speed is $54\ \text{km/h} = 15\ \text{m s}^{-1}$ both before and after, so the size of the momentum, $mv = 0.15\times15 = 2.25\ \text{kg m s}^{-1}$, does not change; only its direction does. The impulse is the vector change in momentum, $\Delta\vec p = m\vec v – m\vec u$, and it is not zero just because the speed is unchanged.

The book reads the question as in the figure: the ball comes in along $AO$ and the batsman sends it away along $OB$, the two paths making an angle of $45^\circ$. Each path then makes $22.5^\circ$ with the bisector of angle $AOB$, so resolve both velocities along and across that bisector. Along the bisector, the incoming ball has a component $15\cos22.5^\circ$ towards the bat and the outgoing ball the same component away from it, so this component reverses. Across the bisector, both have a component $15\sin22.5^\circ$ in the same sense, so it is unchanged. The whole change is therefore along the bisector, away from the bat, and its size is

$$J = 2mv\cos22.5^\circ = 2\times0.15\times15\times0.924 = 4.16\ \text{kg m s}^{-1} \approx 4.2\ \text{kg m s}^{-1}$$

Note on the wording. “Deflects by $45^\circ$” can also be read as the ball’s direction of travel turning through $45^\circ$, a glance off the bat rather than a return. The same method then gives $J = 2mv\sin22.5^\circ \approx 1.7\ \text{kg m s}^{-1}$. The book’s answer key uses the first reading, and so does the answer here.

Question 4.21

A stone of mass $0.25\ \text{kg}$ tied to the end of a string is whirled round in a circle of radius $1.5\ \text{m}$ with a speed of $40\ \text{rev./min}$ in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of $200\ \text{N}$?

Solution. The stone moves in a circle, so the tension must supply the centripetal force $mv^2/R$. First turn the rate of rotation into a speed. Each revolution covers $2\pi R$, and the stone makes $40/60$ revolutions each second:

$$v = 2\pi R\times\frac{40}{60} = 2\pi\times1.5\times\frac23 = 2\pi\ \text{m s}^{-1} \approx 6.28\ \text{m s}^{-1}$$

$$T = \frac{mv^2}{R} = \frac{0.25\times(2\pi)^2}{1.5} = 6.6\ \text{N}$$

The string breaks once $mv^2/R$ would have to exceed $200\ \text{N}$, so the largest speed is

$$v_{\max} = \sqrt{\frac{T_{\max}R}{m}} = \sqrt{\frac{200\times1.5}{0.25}} = \sqrt{1200} \approx 35\ \text{m s}^{-1}$$

Like the book, this treats the string as horizontal and ignores the stone’s weight. Strictly, the string sags a little so that its tension also has an upward component to hold up the $2.5\ \text{N}$ weight, which makes $T$ about $7.0\ \text{N}$ at $40\ \text{rev/min}$. Near $v_{\max}$ the weight is negligible beside $200\ \text{N}$.

Question 4.22

If, in Exercise 4.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks:

Solution. While the string is intact, its tension is the only horizontal force on the stone, and it keeps turning the stone’s velocity towards the centre. The instant the string breaks, that force disappears, and by the first law the stone carries on with the velocity it had at that instant. In circular motion that velocity is along the tangent. Nothing pushes the stone outwards, so (a) is wrong. The outward pull felt when whirling a stone is only the stone’s tendency to go straight on. Once free, gravity then curves its path downwards, but it leaves along the tangent.

Question 4.23

Explain why (a) a horse cannot pull a cart and run in empty space, (b) passengers are thrown forward from their seats when a speeding bus stops suddenly, (c) it is easier to pull a lawn mower than to push it, (d) a cricketer moves his hands backwards while holding a catch.

Question 4.23 (a)

Solution. The horse and the cart pull on each other with equal and opposite forces, by the third law. Taken as one system these forces are internal and cancel, so they cannot set the system moving. On the ground, the horse gets going by pushing backwards on the ground with its hooves. The ground pushes the horse forwards with an equal friction force, and that is an external force on the horse and cart together. In empty space there is nothing to push against, so there is no external force.

Question 4.23 (b)

Solution. When the bus stops, friction from the seat and floor stops the lower part of each passenger’s body along with it. The upper body is not held in the same way, so by its inertia it keeps moving forwards at the bus’s earlier speed until the muscles or the seat in front bring it to rest.

Question 4.23 (c)

Solution. The handle of a lawn mower is at an angle $\theta$ to the horizontal, so the force $F$ applied along it has a vertical component $F\sin\theta$ as well as a horizontal one. When you push, the handle points down into the mower and this component presses it into the ground. For vertical equilibrium the ground must then push up harder: $N = W + F\sin\theta$. When you pull, the component points upwards and partly lifts the mower: $N = W – F\sin\theta$. Friction is proportional to the normal force, so pushing increases the friction to be overcome and pulling reduces it. Pulling therefore needs the smaller force.

Question 4.23 (d)

Solution. To stop the ball, the cricketer must remove its momentum, and that change $\Delta p$ is the same however he does it. The average force on his hands is $F = \Delta p/\Delta t$. Drawing the hands back with the ball stretches out the stopping time $\Delta t$, which reduces the force, and so the sting in his hands.

Common mistakes

• Question 4.2(c): putting the net force at zero at the highest point. The pebble is momentarily at rest there, and students equate rest with no force. But zero velocity is not zero acceleration: the $0.5\ \text{N}$ weight is still acting, which is exactly why the pebble starts to fall again.
• Question 4.3(c): giving the dropped stone the train’s forward push. Once released, the stone is not touching the train, so the train can exert no force on it. It keeps the train’s velocity, not its acceleration, and the net force is its weight alone.
• Question 4.4: choosing $T – mv^2/l$ or $T + mv^2/l$. These treat the centripetal force as a separate force acting alongside the tension. It is not a kind of force at all, only a name for whatever real force points to the centre, and here that force is the tension.
• Question 4.10: using $x = ut + \tfrac12at^2$ right up to $t = 100\ \text{s}$. The formula assumes the force acts throughout, but it stops at $30\ \text{s}$. Applied anyway, it gives $-99\ \text{km}$ instead of $-50\ \text{km}$.
• Question 4.13(a): expecting a higher reading because the lift is going up. The reading depends on the acceleration, not the velocity. At a uniform speed, up or down, the scale reads the man’s true weight.
• Question 4.15: taking the tension to be the same in both cases, or equal to $F$. The string only has to accelerate the body behind it, so the tension is that body’s mass times $a$: $400\ \text{N}$ when the light body is pulled and $200\ \text{N}$ when the heavy one is.
• Question 4.18: answering $0.3\ \text{kg m s}^{-1}$, or zero. Because the speed is unchanged it looks as if nothing has happened. But the velocity reverses, so the momentum goes from $+mv$ to $-mv$, a change of $2mv$.
• Question 4.22: choosing “radially outwards”. The outward pull felt while whirling a stone suggests an outward force. There is none; when the string breaks the stone simply keeps the velocity it had, and that velocity is along the tangent.

Practise next

• Chapter 5, Work, Energy and Power — the same forces, followed through the work they do and the energy they transfer.
• Chapter 3, Motion in a Plane — the projectile paths of 4.11 and 4.12 and the circular motion of 4.21 all rest on it.