Standard +0.3 This is a standard Further Maths mechanics proof involving conservation of energy and the coefficient of restitution formula. Students need to find velocity before impact using mgh = ½mv², apply e = v_after/v_before, then calculate the energy difference. While it requires multiple steps and careful algebraic manipulation, it follows a well-practiced template that Further Maths students encounter regularly in restitution problems.
A ball of mass \(m\) kg is held at rest at a height \(h\) metres above a horizontal surface.
The ball is released and bounces on the surface.
The coefficient of restitution between the ball and the surface is \(e\)
Prove that the kinetic energy lost during the first bounce is given by
$$mgh(1 - e^2)$$
[4 marks]
Question 6:
6 | Finds impact speed or speed2 or
KE in terms of h | 3.3 | B1 | Let U = Impact speed
U 2 = 0 2 + 2 g h
U = 2 g h
Let V = Rebound speed
V =e 2gh
1 1
K E L o s t = m U 2 − m V 2
2 2
= m g h − m g h e 2
= m g h ( 1 − e 2 )
Uses the law of restitution to find
rebound speed in terms of h or
impact speed based on their impact
speed | 1.1a | M1
Forms expression for the KE lost
using their impact and rebound
speeds | 1.1a | M1
Completes a rigorous argument to
show the required result | 2.1 | R1
Total | 4
Q | Marking Instructions | AO | Marks | Typical Solution
A ball of mass $m$ kg is held at rest at a height $h$ metres above a horizontal surface.
The ball is released and bounces on the surface.
The coefficient of restitution between the ball and the surface is $e$
Prove that the kinetic energy lost during the first bounce is given by
$$mgh(1 - e^2)$$
[4 marks]
\hfill \mbox{\textit{AQA Further Paper 3 Mechanics 2021 Q6 [4]}}