(a) Distinguish between perfectly elastic... | WAEC SSCE - Physics - 2010 (Essay)

Question 1 Report

(a) Distinguish between perfectly elastic collision and perfectly inelastic collision.

(b) Sketch a distance — time graph for a particle moving in a straight line with:

(i) uniform speed;

(ii) variable speed.

(c) A body starts from rest and travels distances of 120, 300 and 180m in successive equal time intervals of 12 s. During each interval the body is uniformly accelerated. (i) Calculate the velocity of the body at the end of each successive time interval.

(ii) Sketch a velocity-time graph for the motion.

Answer

(a) Perfectly elastic and perfectly inelastic collisions

Perfectly elastic collision	Perfectly inelastic collision
Both total linear momentum and total kinetic energy are conserved.	Total linear momentum is conserved, but kinetic energy decreases.
The objects separate after the collision.	The objects stick together after the collision and move with a common velocity.
No kinetic energy is converted into other forms such as heat, sound, or deformation.	This is the collision in which the maximum possible kinetic energy is converted into other forms.

(b) Distance–time graphs

The gradient of a distance–time graph represents speed. Therefore, uniform speed gives a constant gradient, whereas variable speed gives a changing gradient.

(c)(i) Velocities at the ends of the 12 s intervals

For uniformly accelerated motion within each interval, the displacement is:

\[s=\frac{(u+v)}{2}t\]

Here, \(t=12\ \text{s}\). The final velocity of one interval is the initial velocity of the next interval.

First interval: \(u=0\), \(s=120\ \text{m}\).

\[120=\frac{(0+v_1)}{2}\times12\]

\[120=6v_1\]

\[v_1=20\ \text{m s}^{-1}\]
Second interval: \(u=20\ \text{m s}^{-1}\), \(s=300\ \text{m}\).

\[300=\frac{(20+v_2)}{2}\times12\]

\[300=6(20+v_2)\]

\[50=20+v_2\]

\[v_2=30\ \text{m s}^{-1}\]
Third interval: \(u=30\ \text{m s}^{-1}\), \(s=180\ \text{m}\).

\[180=\frac{(30+v_3)}{2}\times12\]

\[30=30+v_3\]

\[v_3=0\ \text{m s}^{-1}\]

Time from start / s	Velocity / m s⁻¹
0	0
12	20
24	30
36	0

(c)(ii) Velocity–time graph

Each 12 s interval has uniform acceleration, so each section of the velocity–time graph is a straight line. The area under each section equals the distance travelled in that interval.

Examination reminder: For successive intervals, do not restart from rest each time. Use the velocity at the end of one interval as the initial velocity for the next.

Read lesson note on Rectilinear Acceleration (WAEC)

Read lesson note on Speed And Velocity (WAEC)

Answer Details