Chapter 4: Laws of Motion

Let the wall be vertical along the $$y$$-axis. Take $$x$$-axis horizontal, pointing into the page (away from the wall on the side where the balls travel). Let $$u$$ be the speed of each ball, $$m$$ the mass.

Case (a) — perpendicular incidence: The ball strikes the wall normally with velocity $$+\hat{x}\, u$$ and rebounds with velocity $$-\hat{x}\, u$$.

Initial momentum: $$(p_x)_i = mu, \ (p_y)_i = 0$$.
Final momentum: $$(p_x)_f = -mu, \ (p_y)_f = 0$$.

$$\Delta p_x = -2mu, \quad \Delta p_y = 0$$

The impulse on the ball (and hence the force on the ball) is along the negative $$x$$-direction. By Newton's third law, the force on the wall by the ball is along the positive $$x$$-direction — i.e. normal to the wall, into the wall.

Case (b) — incidence at $$30°$$ with the normal: The initial velocity makes an angle of $$30°$$ with the $$x$$-axis (the normal to the wall).

$$(p_x)_i = mu\cos 30°, \quad (p_y)_i = -mu\sin 30°$$

After reflection (elastic, same speed, angle of reflection = angle of incidence):

$$(p_x)_f = -mu\cos 30°, \quad (p_y)_f = -mu\sin 30°$$

Note that the $$y$$-component (tangential to wall) is unchanged, while the $$x$$-component (normal to wall) reverses.

$$\Delta p_x = -2mu\cos 30°, \quad \Delta p_y = 0$$

So the impulse on the ball is again along the negative $$x$$-direction (normal to wall). By Newton's third law, the force on the wall by the ball is again along the positive $$x$$-direction — normal to the wall, into the wall.

Hence in both cases the force on the wall is directed along the normal to the wall, into the wall. The common, instinctive answer that the force in case (b) is along the line of the incident ball is wrong.

From part (i):

Magnitude of impulse on ball (a): $$|\Delta p|_a = 2mu$$.

Magnitude of impulse on ball (b): $$|\Delta p|_b = 2mu\cos 30°$$.

Therefore the required ratio is:

$$\frac{|\Delta p|_a}{|\Delta p|_b} = \frac{2mu}{2mu\cos 30°} = \frac{1}{\cos 30°} = \frac{2}{\sqrt{3}} \approx 1.2$$

At the optimum speed, the horizontal component of the normal reaction alone provides the centripetal force; the friction force is not called upon. This minimises tyre wear.

For a road banked at angle $$\theta$$, the optimum speed is:

$$v_o = \sqrt{Rg \tan\theta}$$

With $$R = 300 \, \mathrm{m}$$, $$\theta = 15°$$, $$g = 9.8 \, \mathrm{m\,s^{-2}}$$, and $$\tan 15° \approx 0.2679$$:

$$v_o = \sqrt{300 \times 9.8 \times 0.2679}$$

$$v_o = \sqrt{787.6} \approx 28.1 \, \mathrm{m\,s^{-1}}$$

At the maximum permissible speed, the friction acts down the slope (resisting the tendency of the car to slide outward) and is at its limiting value $$f = \mu_s N$$. The maximum speed for a banked road is:

$$v_{\max} = \sqrt{\, R g \cdot \dfrac{\mu_s + \tan\theta}{1 - \mu_s \tan\theta}\,}$$

With $$R = 300 \, \mathrm{m}$$, $$g = 9.8 \, \mathrm{m\,s^{-2}}$$, $$\mu_s = 0.2$$ and $$\tan 15° \approx 0.2679$$:

$$\frac{\mu_s + \tan\theta}{1 - \mu_s\tan\theta} = \frac{0.2 + 0.2679}{1 - 0.2 \times 0.2679} = \frac{0.4679}{0.9464} \approx 0.4944$$

$$v_{\max} = \sqrt{300 \times 9.8 \times 0.4944} = \sqrt{1453.5} \approx 38.1 \, \mathrm{m\,s^{-1}}$$

Before the floor yields, the block is at rest on the floor (only the block, not yet the cylinder — i.e., the block alone). Two forces act on the block:

• weight, $$W = mg = 2 \times 10 = 20 \, \mathrm{N}$$, acting vertically downward, and
• normal reaction, $$R$$, from the floor, acting vertically upward.

By Newton's first law (block at rest):

$$R - W = 0 \implies R = 20 \, \mathrm{N}$$

By Newton's third law, the force on the floor by the block (the action of the block on the floor) is equal and opposite to $$R$$:

$$\text{Force on floor by block} = 20 \, \mathrm{N} \text{ vertically downward.}$$

After the floor yields, treat the block plus the cylinder as one system, total mass:

$$M = m_{\text{block}} + m_{\text{cyl}} = 2 + 25 = 27 \, \mathrm{kg}$$

The system goes down with acceleration $$a = 0.1 \, \mathrm{m\,s^{-2}}$$. Two external forces act on the system: the weight of the system $$Mg$$ acting downward, and the normal reaction $$R'$$ from the floor acting upward.

$$Mg = 27 \times 10 = 270 \, \mathrm{N}$$

Newton's second law (taking downward positive, since the system accelerates downward):

$$Mg - R' = Ma$$

$$270 - R' = 27 \times 0.1 = 2.7$$

$$R' = 267.3 \, \mathrm{N}$$

By Newton's third law, the action of the block on the floor is equal and opposite to $$R'$$:

$$\text{Force on floor by block} = 267.3 \, \mathrm{N} \text{ vertically downward.}$$

Action–reaction pairs:

For case (a):

• The force of gravity on the block by the Earth ($$20 \, \mathrm{N}$$ downward) and the equal and opposite gravitational force on the Earth by the block ($$20 \, \mathrm{N}$$ upward on the Earth).
• The force on the floor by the block ($$20 \, \mathrm{N}$$ downward) and the normal force on the block by the floor ($$20 \, \mathrm{N}$$ upward).

For case (b):

• The gravitational force on the system by the Earth ($$270 \, \mathrm{N}$$ downward) and the gravitational force on the Earth by the system ($$270 \, \mathrm{N}$$ upward on the Earth).
• The force on the floor by the block ($$267.3 \, \mathrm{N}$$ downward) and the normal force on the block by the floor ($$267.3 \, \mathrm{N}$$ upward).
• Internal action–reaction pair (within the system): the force on the block by the cylinder (downward) and the force on the cylinder by the block (upward) — these are internal to the system and so do not appear in its free-body diagram.

Note: the weight of the block (or the system) and the normal reaction of the floor are not action–reaction pairs — they act on the same body and are not of the same nature.

The raindrop is falling with constant speed, hence its velocity is constant — acceleration is zero. By Newton's second law:

$$F_{\text{net}} = ma = 0$$

(The downward weight of the drop is exactly balanced by the upward viscous drag of air and the buoyancy.)

A cork floating on water is in equilibrium — it is at rest, so its acceleration is zero. By Newton's second law, the net force on it is zero.

$$F_{\text{net}} = ma = 0$$

(The weight of the cork acting downward is exactly balanced by the upward buoyant force from the water.)

The kite is held stationary, so it is at rest and its acceleration is zero. By Newton's second law:

$$F_{\text{net}} = ma = 0$$

(Its weight, the tension in the string, and the aerodynamic force due to the wind add to zero.)

The car moves with constant velocity, so its acceleration is zero. By Newton's second law:

$$F_{\text{net}} = ma = 0$$

(The driving force from the engine through the tyres is exactly balanced by the friction/drag forces opposing motion, and the weight is balanced by the normal reaction.)

There is no electric or magnetic field, no gravity from nearby bodies, and no contact force. So no external force acts on the electron.

$$F_{\text{net}} = 0$$

By Newton's first law, the electron continues to move with the same high speed in a straight line.

Air resistance is ignored, so the only force on the pebble at any instant in its motion is its weight, acting vertically downward.

$$F = mg = 0.05 \times 10 = 0.5 \, \mathrm{N}$$

So during upward motion the net force on the pebble is $$0.5 \, \mathrm{N}$$ directed vertically downward.

Again, the only force acting on the pebble is its weight (air resistance is ignored).

$$F = mg = 0.05 \times 10 = 0.5 \, \mathrm{N}$$

Direction: vertically downward.

At the highest point the pebble is momentarily at rest (zero velocity), but the only force acting on it is still gravity. "Momentarily at rest" does not mean zero acceleration — the velocity is changing direction even though its magnitude is instantaneously zero.

$$F = mg = 0.05 \times 10 = 0.5 \, \mathrm{N}, \text{ vertically downward.}$$

If the pebble were thrown at $$45°$$ instead of vertically: The trajectory becomes a parabola, but the only force acting on it (with air resistance neglected) is still its weight. So in every case (a), (b), (c) the net force is the same — $$0.5 \, \mathrm{N}$$ vertically downward.

The one nuance is that at the "highest point" of the $$45°$$ projectile, the pebble is not momentarily at rest — only its vertical component of velocity is zero, while its horizontal component is non-zero. But the net force is unchanged: still its weight, vertically downward.

The stone is in free fall (air resistance neglected). The only force acting on it is its weight.

$$F = mg = 0.1 \times 10 = 1 \, \mathrm{N}$$

Direction: vertically downward.

Since the train moves with constant velocity, the ground frame is an inertial frame and so is the train frame. Once the stone is released, it is no longer in contact with the train; the only force acting on it (air resistance neglected) is its weight.

$$F = mg = 0.1 \times 10 = 1 \, \mathrm{N}, \text{ vertically downward.}$$

(The stone retains the horizontal velocity of the train at the moment of release, but no horizontal force acts on it afterwards.)

The moment the stone is dropped it loses contact with the train; no horizontal force acts on it (air resistance neglected). The acceleration of the train is irrelevant to forces acting on the stone once it is released.

$$F = mg = 0.1 \times 10 = 1 \, \mathrm{N}, \text{ vertically downward.}$$

(Relative to the accelerating train the stone appears to accelerate horizontally backward, but that is a pseudo-effect of viewing motion from a non-inertial frame, not a real force.)

The stone is at rest relative to the train, so in the ground (inertial) frame it has the same acceleration as the train, $$a = 1 \, \mathrm{m\,s^{-2}}$$, in the horizontal direction (the direction the train is accelerating).

The vertical forces — its weight downward and the normal reaction from the floor upward — cancel (no vertical acceleration). The horizontal net force is given by Newton's second law:

$$F_{\text{net}} = ma = 0.1 \times 1 = 0.1 \, \mathrm{N}$$

This horizontal force is provided by static friction from the floor of the train on the stone, in the direction of acceleration of the train.

Up to $$t = 10 \, \mathrm{s}$$, the stone moves with the truck. The horizontal velocity of the truck (starting from rest, accelerating at $$2 \, \mathrm{m\,s^{-2}}$$) just before the stone is dropped is:

$$v_x(10) = 0 + 2 \times 10 = 20 \, \mathrm{m\,s^{-1}}$$ (along the direction of motion of the truck)

At the instant of release ($$t = 10 \, \mathrm{s}$$), the stone has this horizontal velocity and zero vertical velocity.

After release, only gravity acts on the stone (air resistance neglected). The horizontal velocity stays $$20 \, \mathrm{m\,s^{-1}}$$; the vertical component grows at $$g = 10 \, \mathrm{m\,s^{-2}}$$.

At $$t = 11 \, \mathrm{s}$$ (i.e., $$1 \, \mathrm{s}$$ after release):

$$v_x = 20 \, \mathrm{m\,s^{-1}}$$ (horizontal, along the truck's direction)

$$v_y = 0 + g \times 1 = 10 \, \mathrm{m\,s^{-1}}$$ (downward)

Magnitude of the velocity:

$$|\vec{v}| = \sqrt{v_x^2 + v_y^2} = \sqrt{400 + 100} = \sqrt{500} \approx 22.36 \, \mathrm{m\,s^{-1}}$$

Direction (angle below the horizontal):

$$\tan\theta = \frac{v_y}{v_x} = \frac{10}{20} = 0.5 \implies \theta \approx 26.57°$$

Once the stone is released, it loses contact with the truck. The only force acting on it (air resistance neglected) is its own weight $$mg$$. The acceleration of the truck has no bearing on the stone after release.

$$\vec{a} = \vec{g} \implies |\vec{a}| = g = 10 \, \mathrm{m\,s^{-2}}, \ \text{vertically downward.}$$

At an extreme position the bob is momentarily at rest — its instantaneous velocity is zero. The moment the string is cut, the only force acting on the bob is its weight, which is vertically downward. With zero initial velocity, the bob will move along the line of this force.

So the bob falls vertically downward in a straight line (free fall) until it hits the floor.

At the mean (lowest) position of the swing, the bob's velocity is horizontal and equal in magnitude to $$1 \, \mathrm{m\,s^{-1}}$$ (the maximum speed of the oscillation). The moment the string is cut, the only force on the bob is its weight (vertically downward), and it has a horizontal initial velocity.

This is exactly the situation of horizontal projectile motion: the bob has a constant horizontal component of velocity ($$1 \, \mathrm{m\,s^{-1}}$$), while its vertical velocity increases under gravity. Eliminating $$t$$ from $$x = v_0 t$$ and $$y = \tfrac{1}{2} g t^2$$ gives $$y = \dfrac{g}{2 v_0^2} x^2$$, a parabola.

So the trajectory is a parabolic path with the initial horizontal direction tangent to the curve at the point where the string was cut.

The reading of the scale is the normal force $$R$$ that the scale exerts on the man (or equivalently, by Newton's third law, the force the man exerts on the scale).

The lift moves with uniform velocity, so the man's acceleration is zero. Newton's second law (taking upward positive):

$$R - mg = 0 \implies R = mg = 70 \times 10 = 700 \, \mathrm{N}$$

The reading equals the man's true weight: $$\dfrac{R}{g} = 70 \, \mathrm{kg}$$.

Take upward as positive. The lift, and hence the man, has acceleration $$a = -5 \, \mathrm{m\,s^{-2}}$$.

Newton's second law on the man:

$$R - mg = ma$$

$$R = m(g + a) = 70 \times (10 - 5) = 70 \times 5 = 350 \, \mathrm{N}$$

Apparent mass on the scale = $$R/g = 35 \, \mathrm{kg}$$.

Take upward as positive. The lift, and hence the man, has acceleration $$a = +5 \, \mathrm{m\,s^{-2}}$$.

Newton's second law:

$$R - mg = ma$$

$$R = m(g + a) = 70 \times (10 + 5) = 70 \times 15 = 1050 \, \mathrm{N}$$

Apparent mass on the scale = $$R/g = 105 \, \mathrm{kg}$$.

In free fall the lift, the scale and the man all accelerate downward at $$g$$. Take upward as positive, so $$a = -g$$.

$$R - mg = ma = -mg \implies R = 0$$

The scale reads zero — the well-known state of apparent weightlessness. Note that the man's true weight ($$mg = 700 \, \mathrm{N}$$) is unchanged; only the normal force from the scale (i.e. the apparent weight) is zero, because both the man and the scale fall together with the same acceleration.

Impulse equals the change in linear momentum. The velocity changes abruptly at $$t = 0$$ and again at $$t = 4 \, \mathrm{s}$$, so there is an impulsive (infinitely brief but finite-impulse) force at each instant.

At $$t = 0$$: velocity jumps from $$0$$ (for $$t < 0$$) to $$0.75 \, \mathrm{m\,s^{-1}}$$ (for $$0 < t < 4$$).

$$J_0 = m\,\Delta v = 4 \times (0.75 - 0) = 3 \, \mathrm{kg\,m\,s^{-1}}$$

(positive, i.e., in the direction of motion).

At $$t = 4 \, \mathrm{s}$$: velocity jumps from $$0.75 \, \mathrm{m\,s^{-1}}$$ down to $$0$$.

$$J_4 = m\,\Delta v = 4 \times (0 - 0.75) = -3 \, \mathrm{kg\,m\,s^{-1}}$$

(magnitude $$3 \, \mathrm{kg\,m\,s^{-1}}$$, directed opposite to the previous motion).

Let $$m_A = 10 \, \mathrm{kg}$$, $$m_B = 20 \, \mathrm{kg}$$. The surface is smooth (no friction). The two bodies are connected by a light inextensible string; they accelerate together with the same magnitude of acceleration $$a$$ along the line of the string.

Whole system:

$$F = (m_A + m_B)\,a$$

$$600 = 30 a \implies a = 20 \, \mathrm{m\,s^{-2}}$$

Tension in the string: The string accelerates body B (the trailing body) at $$a$$. The only horizontal force on B is the tension $$T$$:

$$T = m_B\,a = 20 \times 20 = 400 \, \mathrm{N}$$

The whole system still has the same acceleration:

$$a = \frac{F}{m_A + m_B} = \frac{600}{30} = 20 \, \mathrm{m\,s^{-2}}$$

Now A is the trailing body (being pulled by the string). The only horizontal force on A is the tension $$T$$:

$$T = m_A\,a = 10 \times 20 = 200 \, \mathrm{N}$$

When a horse pulls a cart on the ground, it pushes its hooves backward and downward against the ground. By Newton's third law, the ground exerts an equal and opposite reaction on the horse — directed forward and upward. The forward component of this reaction is what propels the horse (and the cart) ahead. This is possible because of the friction between the hooves and the ground.

In empty space there is no ground (and no other body) to push against. The horse can push its legs as much as it likes — there is no surface to react back on it. Without that reaction force, the horse has no external horizontal force, and so cannot accelerate or run.

This is an example of inertia of motion (Newton's first law).

While the bus is moving, the passengers are also moving forward with the same velocity. When the bus suddenly brakes, the brakes apply a large retarding force on the bus, but no comparable external force acts directly on the upper body of the passenger. By Newton's first law, the passenger's upper body tends to continue moving forward with the original velocity — and so it is thrown forward relative to the (now decelerating) seat.

The friction between seat and the passenger's lower body can decelerate the lower body to some extent, but the upper body — not in firm contact — continues forward, giving the felt sensation of being thrown forward.

Let $$F$$ be the force applied along the handle of the lawn mower, which makes some angle $$\theta$$ above the horizontal. The mower has weight $$mg$$, normal reaction $$N$$ from the ground, and friction $$f = \mu N$$ opposing motion.

Pulling: the applied force has a horizontal component $$F\cos\theta$$ (forward) and a vertical component $$F\sin\theta$$ directed upward. The vertical balance gives:

$$N_{\text{pull}} = mg - F\sin\theta$$

So $$N_{\text{pull}} < mg$$, and the frictional force $$\mu N_{\text{pull}}$$ is reduced.

Pushing: the applied force has horizontal component $$F\cos\theta$$ (forward) and vertical component $$F\sin\theta$$ now directed downward. Hence:

$$N_{\text{push}} = mg + F\sin\theta$$

So $$N_{\text{push}} > mg$$, and the frictional force is increased.

Since $$f_{\text{pull}} < f_{\text{push}}$$, less force is needed to move the lawn mower when pulling than when pushing — pulling is easier.

By Newton's second law in its impulse form:

$$F = \frac{\Delta p}{\Delta t}$$

The change in momentum $$\Delta p$$ of the ball during the catch is fixed (it goes from its incoming momentum down to zero). By moving his hands backward, the cricketer increases the time $$\Delta t$$ over which the ball is brought to rest. Since $$F \propto 1/\Delta t$$, the average force he has to apply on the ball — and therefore the force the ball applies on his hand by Newton's third law — is reduced.

This protects his hand from injury, by spreading the same change of momentum over a longer time and thus lowering the peak force.