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mememe1245

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**Question**

A 0.0420-kg hollow racquetball with an initial speed of 12.0 m/s collides with a backboard. It rebounds with a speed of 6.0 m/s.

a. Calculate the total impulse on the ball.

b. If the contact time lasts for 0.040 s, calculate the average force on the ball.

**What I did on the test**:

(A)Impulse=F x change in T=change in momentum(mass x velocity)

F= unknown, so use the formula, F=M x A

_________________________

Momentum=

P(momentum)= 0.0420kg x 6m/s

P= .252N

Therefore, the impulse on the ball is, .252kg. m/s.

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(B) F= Mass x Acceleration

change in Velocity= Acceleration x Time

Acceleration= change in Velocity/ Time

A= Velocity final - Velocity initial/ Time

A= 6m/s -12m/s/0.040s

A= -6m/s/0.040s

A= - 150 m/s^2

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F= M x A

F= 0.042kg x (-150m/s^2)

F= -6.3 N

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