(a) Distinguish between perfectly elastic collision and perfectly inelastic collision. (b) Sketch a distance — time graph for a particle moving in a straigh...

Answer Details

(a) Perfectly elastic vs perfectly inelastic collision

Perfectly elastic collision	Perfectly inelastic collision
Both momentum and kinetic energy are conserved.	Momentum is conserved but kinetic energy is not; some is converted to heat, sound and deformation.
The colliding bodies separate (do not stick) after impact.	The colliding bodies stick together and move as one body after impact.
The relative velocity of separation equals the relative velocity of approach.	The bodies move with a single common velocity after impact (relative velocity of separation = 0).

(b) Distance–time graphs

For uniform speed the distance increases by equal amounts in equal times, so the graph is a straight line of constant gradient through the origin. For variable speed the speed keeps changing, so the gradient keeps changing and the graph is a curve (here the gradient increases, showing increasing speed).

(c) Body starting from rest

The body covers 120 m, 300 m and 180 m in three successive equal intervals of \(t = 12\ \text{s}\), being uniformly accelerated within each interval. For uniform acceleration the average velocity over an interval equals the mean of the velocities at its two ends, so the end-of-interval velocities follow directly.

(i) Velocity at the end of each interval

Interval 1 (\(u = 0\)): \(s = ut + \tfrac{1}{2}at^2\)

\[120 = 0\times 12 + \tfrac{1}{2}a(12)^2 = 72a \;\Rightarrow\; a = \tfrac{5}{3}\ \text{m s}^{-2}\] \[v_1 = u + at = 0 + \tfrac{5}{3}\times 12 = 20\ \text{m s}^{-1}\]

Interval 2 (\(u = v_1 = 20\ \text{m s}^{-1}\)):

\[300 = 20\times 12 + \tfrac{1}{2}a(12)^2 = 240 + 72a \;\Rightarrow\; a = \tfrac{5}{6}\ \text{m s}^{-2}\] \[v_2 = 20 + \tfrac{5}{6}\times 12 = 30\ \text{m s}^{-1}\]

Interval 3 (\(u = v_2 = 30\ \text{m s}^{-1}\)):

\[180 = 30\times 12 + \tfrac{1}{2}a(12)^2 = 360 + 72a \;\Rightarrow\; a = -2.5\ \text{m s}^{-2}\] \[v_3 = 30 + (-2.5)\times 12 = 0\ \text{m s}^{-1}\]

So the velocities at the end of the successive intervals are 20 m s\(^{-1}\), 30 m s\(^{-1}\) and 0 m s\(^{-1}\) (at \(t = 12\ \text{s}, 24\ \text{s}\) and \(36\ \text{s}\) respectively). Each result checks against the average speed: \(120/12 = 10 = \tfrac{0+20}{2}\), \(300/12 = 25 = \tfrac{20+30}{2}\), \(180/12 = 15 = \tfrac{30+0}{2}\).

(ii) Velocity–time graph

The graph is made of three straight segments (constant acceleration in each interval): rising from 0 to 20 m s\(^{-1}\) in the first 12 s, rising more gently to 30 m s\(^{-1}\) at 24 s, then falling to rest at 36 s.