A 0.620-kg ball is dropped from rest at a point 4.40 m above the floor. The ball

Question

A 0.620-kg ball is dropped from rest at a point 4.40 m above the floor. The ball rebounds straight upward to a height of 2.00 m. Taking the negative direction to be downward, what is the impulse of the net force applied to the ball during the collision with the floor?

Hint given:
We need to calculate the velocity of the ball just before and just after it rebounds from the floor and apply equation 7.4 to calculate the impulse.
The velocity of the ball just before and just after the collision can be calculated two ways;
(1) By conservation of energy, .5(m*vf)^2+m*g*hf=.5(m*vo)^2+m*g*ho
(2) Using kinematics, (v of y)^2=(v of oy)^2+2ayHint given:
We need to calculate the velocity of the ball just before and just after it rebounds from the floor and apply equation 7.4 to calculate the impulse.
The velocity of the ball just before and just after the collision can be calculated two ways;
(1) By conservation of energy, .5(m*vf)^2+m*g*hf=.5(m*vo)^2+m*g*ho
(2) Using kinematics, (v of y)^2=(v of oy)^2+2ay

Explanation / Answer

First we find the velocity of the ball right when it hits the floor by using the conservation of energy: mgh = (1/2)mv^2 2gh = v^2 v = sqrt(2gh) = sqrt(2(9.8)(4.40)) = -9.29 m/s (since it is moving downwards) Then using the height to which the ball rises again we can find the velocity right after it hits the floor by conserving energy again: (1/2)mv^2 = mgh v = sqrt(2gh) = sqrt(2*9.8*2.00) = 6.26 m/s Now we know that impulse is equal to the change in momentum Fdt = mdv Impulse = mdv = m(vfinal - vinitial) = m(6.26 - (-9.29)) = (0.620 kg)(15.55 m/s) = 9.64 Ns

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