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The Question:

Steel balls of mass=0.10kg are dropped from a height of ℎ=5.0m onto a smooth steel plate inclined at 45∘ to the vertical.

If the balls are dropped one at a time at a rate of n=100 s^-1, what is the average force on the plate? You may assume that the collisions between the ball and the plate are elastic and that each ball hits the plate only once.

My Problem:

I understand that all of the Gravitational potential energy is converted to kinetic energy mgh = 0.5mv^2 where v turns out to be 10g where g is 9.81ms^-2. But I cannot get the force exerted by the plate using Force x Time = Momentum

Any help would be appreciated

Thank you

Anthony Cheng
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1 Answers1

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Notice, vertical velocity of each ball achieved by dropping a vertical height $h$ $$v=\sqrt{2gh}=\sqrt{2\times 9.81\times 5}=\sqrt{98.1}$$ Now the component of vertical velocity $v$ at $45^\circ$ to the plane which is $v\cos 45^\circ$ will be normal to the inclined plane & hence it will reverse in the direction after collision leaving no change in the magnitude.

There will be no change in the momentum parallel to the inclined plane.

Hence, the change in the momentum of each ball (mass, $m=0.1\ kg$) in the perpendicular direction to the plane $$m(v\cos 45^\circ-(-v\cos 45^\circ))$$ $$=2mv\cos 45^\circ=\sqrt 2 mv$$ Now, the force exerted by the balls on the plane will be equal to the total rate of change of the momentum of the balls $$=\text{(change in momentum of each ball)}\times \text{(number of balls hitting per second)}$$ $$=\sqrt 2 mv\times 100$$ $$100\sqrt 2\times 0.1\times \sqrt {98.1}$$$$=10\sqrt{196.2}\approx 140.07\ N\ \text{(Newtons)}$$

Harish Chandra Rajpoot
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