Exam practice

They have different masses, so different inertia. A more massive vehicle has more inertia of motion, and for a given braking force its retardation is smaller (a = F/m), so it takes longer to stop. A train, with an enormous mass, takes the longest; a motorcycle the shortest.

Because of inertia of rest. The shaking suddenly moves the branches, but the leaves and fruits tend to stay at rest. The stalks are pulled hard, snap, and the leaves and fruits fall.

Because of inertia of motion. The bag was moving forward with the bus. When the bus slows suddenly, the floor’s small friction can’t slow the bag as quickly, so the bag keeps moving forward and slides along the floor.

By Newton’s third law, when the gun pushes the bullet forward (action), the bullet pushes the gun backward with an equal force (reaction): the gun recoils. Held loosely, it would jerk back hard and could hurt the shooter. Held firmly against the shoulder, the recoil is shared with the much larger mass of the body, so it moves back only a little (a = F/m).

They have different elasticity (different rubber, or different air pressure inside). On hitting the floor, each ball is squashed and springs back. The more elastic ball returns more of the energy of its fall as it regains its shape, and wastes less as heat, so it bounces higher.

A rubber band regains its shape only while it is stretched within its elastic limit. Stretched beyond it, it is permanently deformed (it stays longer and loose), and stretched further still, it snaps.

Average velocity is the total displacement of a body divided by the total time taken: average velocity = total displacement ÷ total time. It has a direction (that of the displacement). For uniform acceleration it also equals $\tfrac{u+v}{2}$.
Acceleration is the rate of change of velocity: the change in velocity per unit time, $a = \dfrac{v-u}{t}$. Its SI unit is m/s².

Let a body moving in a straight line with uniform acceleration a change its velocity from u to v in time t, covering a displacement s.

• a.By definition, $a = \dfrac{v-u}{t}$. So $at = v - u$, and $v = u + at$.
• b.For uniform acceleration, average velocity = $\dfrac{u+v}{2}$, so $s = \left(\dfrac{u+v}{2}\right) t$. From (a), $t = \dfrac{v-u}{a}$. Then

$s = \dfrac{(v+u)}{2} \times \dfrac{(v-u)}{a} = \dfrac{v^2-u^2}{2a}$, so $2as = v^2 - u^2$, and $v^2 = u^2 + 2as$.
c. From $s = \left(\dfrac{u+v}{2}\right) t$ and $v = u + at$:

$s = \left(\dfrac{u + u + at}{2}\right) t = \left(u + \tfrac{1}{2}at\right) t$, so $s = ut + \tfrac{1}{2}at^2$.

Displacement–time graph: a straight line sloping upward (starting from the origin if the body starts at the reference point). Its slope is the uniform velocity.

Velocity–time graph: a horizontal straight line, parallel to the time axis, at the height of the velocity. Its slope is zero: no acceleration.

They are the second picture in Figure 7.8 and the first in Figure 7.10.

0–4 s: a straight line rising; uniform velocity forward. 4–6 s: flat: at rest at 8 m. 6–8 s: straight, rising: uniform velocity forward again (4 m in 2 s = 2 m/s). 8–14 s: straight, falling: it comes back to the start with uniform velocity (−12 m in 6 s = −2 m/s).
Average velocity in the first 4 s = slope = $\dfrac{8 - 0}{4 - 0}$ = 2 m/s forward.

The hare (A to C) sets off very fast: in 3 minutes it is 60 m along the 100 m track, far ahead of the tortoise. Sure of winning, it lies down for a nap: from C to D its line is flat, so it is at rest for 7 minutes. Meanwhile the tortoise’s straight line climbs steadily: it walks at the same slow, uniform speed all the way. At about 7 minutes the tortoise’s line crosses the hare’s: it passes the sleeping hare. At D (10 minutes) the hare wakes and runs faster and faster (D to E to F, a curve getting steeper), but at 12 minutes the tortoise reaches the finish, B, while the hare, at F, is still a few meters short. Slow and steady wins the race.

Inertia is the property of a body by which it cannot by itself change its state of rest or of uniform motion in a straight line.

Inertia of rest: (1) a passenger falls backward when a bus starts suddenly; (2) dust comes out of a carpet when it is beaten with a stick.

Inertia of motion: (1) a passenger falls forward when a moving bus brakes suddenly; (2) a long-jumper runs before jumping so as to jump farther.

The inertia of a body depends on its mass: the greater the mass, the greater the inertia, and the smaller the mass, the smaller the inertia. Mass is a measure of inertia. For example, a loaded truck is much harder to start or stop than an empty one.

A body at rest remains at rest, and a body in motion continues to move in a straight line with uniform velocity, unless an external unbalanced force acts on it.

• 1.A footballer kicks a ball at rest: the force from the foot sets it moving.
• 2.The brakes of a bicycle and the friction of the road bring a moving bicycle to rest.
• 3.A batsman’s hit changes the direction of a cricket ball.

By Newton’s first law, a moving bus tends to keep going in a straight line. To go round a bend, it needs a sideways unbalanced force, which comes from the friction of the road on its tyres. The faster the bus goes, the larger this force must be (and the sharper the bend, the larger still). On a sharp mountain bend at high speed, the force needed can be more than the friction can give, especially on a wet, gravelly or muddy road. The bus then carries on almost straight, off the edge of the road. Its large mass (inertia) also means it takes a long distance to slow down, and passengers are thrown sideways. So buses must go slowly on winding roads.

Second law: the acceleration produced in a body is directly proportional to the force acting on it and inversely proportional to its mass.

Proof: $a \propto F$ (mass constant) and $a \propto \dfrac{1}{m}$ (force constant). Combining them, $a \propto \dfrac{F}{m}$, so $F \propto ma$ and $F = k\,ma$, where k is a constant.

One newton is defined as the force that gives a 1 kg mass an acceleration of 1 m/s². Putting F = 1 N, m = 1 kg, a = 1 m/s²: 1 = k × 1 × 1, so k = 1. Therefore F = ma.

To every action there is an equal and opposite reaction. If a body A exerts a force on a body B, then B exerts a force on A that is equal in magnitude and opposite in direction. The two forces act on different bodies.

First law: (1) passengers fall forward when a bus brakes suddenly; (2) a carpet is cleaned by beating it.

Second law: (1) an empty cart is easier to accelerate than a loaded one with the same push; (2) a cricket ball hit harder goes faster.

Third law: (1) we walk by pushing the ground backward, and the ground pushes us forward; (2) a gun recoils when it fires a bullet.

• a.Action: the racket pushes the ball forward. Reaction: the ball pushes the racket backward (you feel it in your hand).
• b.Action: the man’s feet push the boat backward. Reaction: the boat pushes the man forward, towards the bank. Because the boat is free to move on water, it moves backward away from the bank, and if the man isn’t careful he falls into the water.

Elasticity is the property of matter by which a body regains its original shape and size when the deforming force is removed. Examples: a rubber band, a steel spring, the strings of a racket, a football.
Plasticity is the property of matter by which a body keeps its deformed shape after the deforming force is removed. Examples: clay, wet mud (bricks), dough, a toothpaste tube.

u = 0, a = 1.5 m/s², t = 30 s.
$s = ut + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2} \times 1.5 \times 30^2 = 0.75 \times 900$ = 675 m.
$v = u + at = 0 + 1.5 \times 30$ = 45 m/s towards the south.

The stone is dropped, so u = 0; t = 2 s; g = 9.8 m/s².
$h = ut + \tfrac{1}{2}gt^2 = 0 + \tfrac{1}{2} \times 9.8 \times 2^2$ = 19.6 m.

The total mass moving is m = 50 + 12 = 62 kg.
F = ma = 62 × 2 = 124 N (down the hill).

u = 72 km/h = 72 × 1000 ÷ 3600 = 20 m/s; v = 10 km/h = 10 × 1000 ÷ 3600 = 2.78 m/s; s = 50 m.
$v^2 = u^2 + 2as \Rightarrow 2.78^2 = 20^2 + 2 \times a \times 50 \Rightarrow 7.72 = 400 + 100a \Rightarrow a = -3.92\ \text{m/s}^2$
F = ma = 1500 × (−3.92) = −5880 N: a braking force of about 5900 N, opposite to the motion.
(If you round the retardation to 3.9 m/s², you get 1500 × 3.9 = 5850 N. Either is right with the working shown.)

For (iv), the displacement is the area under the velocity–time line between C and D.

• i.Uniform velocity: BC and DE (flat). Uniform acceleration: AB, CD and EF (straight, sloping up). Retardation: FG; its line is curved, so it is a variable retardation (the velocity falls more and more quickly). The only segment where the acceleration is variable is this curved one, FG.
• ii.The steepest of AB (10 m/s in 4 s = 2.5 m/s²), CD (15 m/s in 4 s = 3.75 m/s²) and EF (5 m/s in 4 s = 1.25 m/s²) is CD.
• iii.$a = \dfrac{v-u}{t} = \dfrac{25-10}{10-6} = \dfrac{15}{4}$ = 3.75 m/s².
• iv.The area under CD is a trapezium with parallel sides 10 m/s and 25 m/s, 4 s wide:

displacement = $\dfrac{10 + 25}{2} \times 4$ = 70 m. (Check: $s = ut + \tfrac{1}{2}at^2 = 10 \times 4 + \tfrac{1}{2} \times 3.75 \times 16 = 40 + 30 = 70$ m.)