a raquet ball with mass m=0.224 kg is moving toward the wall at v=15 m/s and at

Question

a raquet ball with mass m=0.224 kg is moving toward the wall at v=15 m/s and at an angle of 27 degrees with respect to the horizontal the ball makes a perfectly elastic collision with the solid frictinless wall and rebounds at the same angle with respect to the horizontal the ball is in contact with the wall t=0.077 s 1). what is the magnitude of the change in momentum of the raquet ball? 2). what is the magnitude of the average force the wall exerts on the raquet ball?

Now the raquet ball is moving straight toward the wall at a velocity of vi=15 m/s the ball makes an inelastic collision with the solid wall and leaves the wall in the opposite direction at vf=9.9 m/s the ball exerts the same average force on the ball as before 3). what is the magnitude of the change in momentum of the raquet ball? 4). what is the time the ball is in contact with the wall?

Explanation / Answer

1) Given mass of raquet ball =0.224kg

Velocity of raquet ball=15m/s

Angle of impact with wall=27 degree (with the horizontal)

Since impact is elastic, the magnitude of velocity will not change after impact. To solve this problem first break velocity vector in component form i.e vcos(27) and vsin(27) along horizontal and vertical respectively. The horizontal component cancel each other, only vertical component remains.

Therefor the magnitude of the change in momentum of the raquet ball = 2 mvsin(27)

=6.42 kgm/s

2) Average force the wall exarts on the raquet ball =dp/dt

=0.224x15xcos(27)/0.077

=12.74N

3) The magnitude of the change in momentum of the raquet ball =mvf - mvi

=0.224x5.1

=1.14kgm/s

4) Time of contact of the ball with wall=dp/average force

=1.14/12.74

=0.089 s

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