AS June 2018 Q1
1. A small ball of mass 0.3 kg is released from rest from a point 3.6 m above horizontal ground. The ball falls freely under gravity, hits the ground and rebounds vertically upwards.
In the first impact with the ground, the ball receives an impulse of magnitude 4.2 N s.
The ball is modelled as a particle.
| Scheme | Marks | AO |
|---|---|---|
| Speed just before impact: \(v^2 = u^2 + 2as = 2 \times 9.8 \times 3.6\ (= 70.56)\) | M1 | 3.4 |
| \(v = 8.4\ (\text{m s}^{-1})\) | A1 | 1.1b |
| Use of \(I = mv - mu\): \(4.2 = 0.3\big(w - (-8.4)\big)\) | M1 | 3.1b |
| Follow their 8.4 | A1ft | 1.1b |
| \(w = 5.6\ (\text{m s}^{-1})\) | A1 | 1.1b |
| (5) |
Notes
M1: Use the model and suvat or energy to find speed before impact
A1: Correct answer. Accept \(\sqrt{70.56}\), \(\sqrt{7.2g}\)
M1: A complete strategy to find \(w\): Use the model and impulse-momentum equation using given impulse and their speed of impact. Must be using a difference in velocities. Be vigilant for sign fudges that make the original equation incorrect.
A1ft: Correct unsimplified equation using their speed
A1: Correct positive answer
| Scheme | Marks | AO |
|---|---|---|
| KE lost \(= \dfrac{1}{2}m\left(v^2 - w^2\right)\) | M1 | 3.3 |
| \(= \dfrac{0.3}{2}\left(8.4^2 - 5.6^2\right)\) Follow their 8.4 and 5.6 | A1ft | 1.1b |
| \(= 5.88\ (\text{J})\) | A1 | 1.1b |
| (3) | ||
| (8 marks) |
Notes
M1: Correct method to find the KE lost in the impact. Need to be using speeds immediately before and immediately after impact.
A1ft: Correct expression for their speeds. Accept subtraction either way round
A1: Correct solution only. Accept 5.9