1.A very hard rubber ball (m=0.3kg) is falling vertically at 9m/s just before it

Question

1.A very hard rubber ball (m=0.3kg) is falling vertically at 9m/s just before it bounces on the floor. The ball rebounds back at essentially the same speed. If the Collison with the floor lasts 0.05s, what is the average force exerted by the floor on the ball?
2.A tennis ball (m=0.2kg) is thrown at a brick wall. It is traveling horizontally at 14 m/s just before hitting the wall and rebounds from the wall at 12m/s, still traveling horizontally.The ball is contact with the wall 0.02s. What is the magnitude of the average force of the wall on the ball?
3. A woman mass of 43kg runs at a speed of 10m/s and jumps onto a giant 30 kg skateboard initially at rest.what is the combined speed of the woman and the skateboard?
4. A 3kg ball traveling to the right with a speed of 4m/s collides with a 5 kg ball traveling to the left with a speed of 5m/s. Take a right to be positive direction. what is the total momentum of the two Balls before they collide? what is the total momentum after the two balls collide?

Explanation / Answer

1)

change in momentum = 2mv = 2*0.3*9 = 5.4 kg m/s.

hence impulse = 5.4 kg m/s which is nothing but the average force times the time taken.

Average force = 5.4/0.05 = 108 N.

2) Applying the same formula as explained above,

average force = change in momentum/ time taken = 0.3*(14+12)/0.02 = 390 N.

3) Initial momentum = mv = 43*10 = 430 kg m/s.

Due to the momentum conservation, the final momentum = 430 kg m/s

hence (m+M)*v_f = 430

73*v_f = 430

v_f = 5.89 m/s

4) The momentum before they collide = m1v1+m2v2 = 3*4-5*5 = -13 kg m/s

Since the momentum is conserved here, the final momentum will be equal to = -13 kg m/s

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