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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.
****************************************************************
(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
_____________
F= M x A
F= 0.042kg x (-150m/s^2)
F= -6.3 N
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.
****************************************************************
(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
_____________
F= M x A
F= 0.042kg x (-150m/s^2)
F= -6.3 N
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