A tennis ball of mass m t is held just above a basketball of mass m b , as shown

Question

A tennis ball of mass mt is held just above a basketball of mass mb, as shown in the figure below. With their centers vertically aligned, both are released from rest at the same moment, so that the bottom of the basketball falls freely through a height h and strikes the floor. Assume an elastic collision with the ground instantaneously reverses the velocity of the basketball while the tennis ball is still moving down because the balls have separated a bit while falling.

A tennis ball of mass mt is held just above a basketball of mass mb, as shown in the figure below. With their centers vertically aligned, both are released from rest at the same moment, so that the bottom of the basketball falls freely through a height h and strikes the floor. Assume an elastic collision with the ground instantaneously reverses the velocity of the basketball while the tennis ball is still moving down because the balls have separated a bit while falling. The two balls meet in an elastic collision. To what height does the tennis ball rebound? (Use any variable or symbol stated above along with the following as necessary: g.) ht = How do you account for the height in (a) being larger than h? Does that seem like a violation of conservation of energy?

Explanation / Answer

initial momentum = mt+mb) *sqrt(2gh)

after collision of basketball with floor,

mb*V -mt*V' = (mt+mb) *sqrt(2gh) --------------------------eqtn1

kinetic energy conservation equation

1/2*(mt+mb)*2gh = 1/2 *mt*V'2 + 1/2*mb*V^2 -----------------eqtn2

2 unknown, V and V'

2 eqtns,

solving

will give V and V'

so velocity of tennnis ball after collsiion with basketball = V'

mt*V' = mt*V2

so rebound velocity will be same as V'

ht = V'2/2g

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