A ball and a lump of plasticine of equal mass are thrown at a wall. The ball is in contact with the wall for 0.05 s it then bounces back. The plasticine hits the wall and doesn't bounce back....

A ball and a lump of plasticine of equal mass are thrown at a wall. The ball is in contact with the wall for 0.05 s it then bounces back. The plasticine hits the wall and doesn't bounce back. Explain why the ball exerts a larger force on the wall then the plasticine.

Asked on by luccamole

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ishpiro | College Teacher | (Level 1) Educator

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The force exerted on the wall will depend on the change of momentum of either ball or plasticine. The change of momentum is related to the force asĀ 

`Delta vecP = vecF* Delta t`

The force in this equation is the force exerted from the wall on the object, but according to the third Newton's Law it is equal in magnitude and opposite in direction to the force acting from the object to the wall. The force is therefore directed away from the wall.

The vector change of momentum equals the difference between the final and initial momentum:

In the case of the ball, the initial momentum is directed towards the wall and, since the ball bounces back, there is a non-zero final momentum, directed away from the wall. Then, the magnitude of the momentum change equals

`DeltaP = P_f + P_i`

In the case of the plasticine, however, the final momentum is zero, since the plasticine gets stuck to the wall. So the magnitude of the momentum change is

`DeltaP = P_i`

The magnitude of the force exerted on the wall by the ball is then

`(P_f+P_i)/(Deltat)`

and the magnitude of the force exerted on the wall by the plasticine is

`P_i/(Deltat)`

Assuming the initial momentum of the ball and plasticine is the same (that is, they are thrown at the wall with the same speed, since their masses are the same), this means that the force exerted on the wall by the ball is greater than the force exerted on the wall by the plasticine.

Sources:

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