4. (a) Which of these is a unit of momentum? A kg/m/s B kg/m/s C kg/m/s² D kg/m/s² (b) Students investigate conservation of momentum using two identical trolleys - Edexcel - GCSE Physics Combined Science - Question 4 - 2022 - Paper 1

To investigate momentum conservation, the students can set up two identical trolleys on a level track. Trolley A is given a card, while Trolley B remains stationary.

1. Setup: Position the two trolleys on a smooth, level runway with sticky pads facing each other to catch the trolleys after they collide. Ensure there are no external forces acting on the trolleys, such as friction.

2. Measurement: Use a ruler to measure the initial velocity of Trolley A before the collision.

3. Collision: Release Trolley A, allowing it to collide with Trolley B. Record the final velocities of both trolleys after the collision using a timing mechanism or video analysis.

4. Analysis: Before the collision, calculate the momentum of Trolley A using the formula: ( p = mv ). After the collision, calculate the momentum of Trolley B. Compare the total momentum before and after to verify momentum conservation.

The runway is sloped to ensure that Trolley A can maintain a constant speed without slowing down due to friction. A downward slope allows gravity to assist in maintaining momentum, making the investigation more effective in demonstrating momentum conservation.

To find the average force exerted by the ball on the racket, we apply the formula:

$F = \frac{mv - mu}{t}$

Where:

• m is the mass of the ball = 0.075 kg
• v is the final velocity (15 m/s, the ball is moving away from the racket)
• u is the initial velocity (8.2 m/s, the ball is moving towards the racket)
• t is the time of contact = 12 ms = 0.012 s

Substituting the values:

$F = \frac{(0.075)(15) - (0.075)(8.2)}{0.012}$

Calculating:

1. Find ( 0.075(15) = 1.125 )
2. Find ( 0.075(8.2) = 0.615 )

Now substitute back into the force equation:

$F = \frac{1.125 - 0.615}{0.012}$ $F = \frac{0.51}{0.012} = 42.5 \, N$

Thus, the average force exerted by the ball on the racket is 42.5 N.

Newton's Third Law states that for every action, there is an equal and opposite reaction. In the context of the collision between the racket and the ball:

1. When the ball strikes the racket, it exerts a force on the racket in the direction of its motion.
2. Simultaneously, the racket exerts an equal and opposite force on the ball, causing it to change direction and travel away from the racket.
3. This exchange of forces exemplifies the principle of action and reaction, confirming that both forces act on different bodies (the racket and the ball).