108k views
4 votes
A 50 kg skater at rest on a frictionless rink throws a 2 kg ball, giving the ball a velocity of 20 m/s. What is the subsequent motion of the skater?

User Ayohaych
by
7.2k points

1 Answer

6 votes

Answer:

Momentum of ball = mass of ball x velocity of ball

P(ball) = 2 kg x 20 m/s = 40 kg*m/s

Step-by-step explanation:

According to the law of conservation of momentum, the total momentum of the system (skater and ball) must remain constant before and after the throw.

Let's first calculate the momentum of the ball:

Momentum of ball = mass of ball x velocity of ball

P(ball) = 2 kg x 20 m/s = 40 kg*m/s

Since the skater was at rest before throwing the ball, the initial momentum of the system was 0. Therefore, the final momentum of the system after the throw must also be 40 kg*m/s to conserve momentum.

The momentum of the skater after the throw can be calculated as follows:

P(skater) = P(system) - P(ball)

P(skater) = 40 kgm/s - (2 kg x 20 m/s)

P(skater) = 0 kgm/s

This means that the skater has no momentum after throwing the ball. Since momentum is equal to mass times velocity, the skater's velocity must also be 0. Therefore, the skater remains at rest on the frictionless rink after throwing the ball.

User KLI
by
8.2k points