A tennis ball of mass 54.0 g is held just above a basketball of mass 598 g. With

Question

A tennis ball of mass 54.0 g is held just above a basketball of mass 598 g. With their centers vertically aligned, both are released from rest at the same moment, to fall through a distance of 1.12 m


(a) Find the magnitude of the downward velocity with which the basketball reaches the ground. Assume an elastic collision with the ground instantaneously reverses the velocity of the basketball while the tennis ball is still moving down.


(b) Next, the two balls meet in an elastic collision. To what height does the tennis ball rebound?

Explanation / Answer

a) we can do this a couple ways. I'll show the energy. The potential energy at beginning is equal to the kinetic at the ground.

PE=KE
mgh=(1/2)mv2

v=(2gh=[2(9.8)1.12]

v=4.69 m/s

As we noticed the mass of the equation doesn't effect the speed 1.12m. So we can assume the tennis ball is moving the same velocity. Also, because the collision of the basketball is elastic we can assume the the speed is 4.69 upward after it rebounds.

So with that we can use the conservation of momentum equation.:

mbvb1+mtvt1=mbvb2+mtvt2

(.598)4.69+.054(-4.69)=(.598)vb2+(.054)vt2

2.55136=(.598)vb2+(.054)vt2

vt2=2.55136-(.598)vb2

Also consevation of energy:

(.652)4.692=(.598)vb22+(.054)vt22

14.3414572=(.598)vb22+(.054)(2.55136-(.598)vb2)2

vb2=-4.8961636

vt2=5.47657 m/s

Finally use the kinematic equation

y=-4.9(v/g)2+v(v/g)

y=1.53 m

Hope that helps

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