A tennis ball of mass 57.0 g is held just above a basketball of mass 637 g. With

Question

A tennis ball of mass 57.0 g is held just above a basketball of mass 637 g. With their centers vertically aligned, both balls are released from rest at the same time, to fall through a distance of 1.35 m, as shown in the figure below. (a) Find the magnitude of the downward velocity with which the basketball reaches the ground. Incorrect: Your answer is incorrect. Your incorrect answer may have resulted from roundoff error. Make sure you keep extra significant figures in intermediate steps of your calculation. m/s (b) Assume that an elastic collision with the ground instantaneously reverses the velocity of the basketball while the tennis ball is still moving down. Next, the two balls meet in an elastic collision. To what height does the tennis ball rebound? m

Explanation / Answer

This problem is equivalent to having the basketball collide head-on with the tennis ball, when they have the same magnitude of velocity (but in opposite directions). The falling to ground just before the collison only serves to give both balls their velocity (same speed), and to reverse the direction of the basketball, so there is a collison. (In a real experiment, the tennis ball would collide with the basketball before the basketball attained full reverse speed. And of course the collisions would be significantly inelastic.)

Therefore set up the problem as any elastic collison problem: Let 1 be basketball and 2 be tennis ball. Let up be the positive direction. Because ball 1 is heavier than ball 2, and because both have same speed, then ball1 will continue upward, at reduced speed, while ball 2 reverses direction. Before the collison the balls have the same speed, same magnitude of velocity, call it v0. Let balls 1 and 2 have velocities v1 and v2 after the collision.

Conservation of momentum:
m1 v0 - m2 v0 = m1 v1 + m2 v2 (Eqn 1)

Conservation of kinetic energy: The coefficient (1/2) is factored out.
m1 (v0)^2 + m2 (v0)^2 = m1 (v1)^2 + m2 (v2)^2 (Eqn 2)

You now have 2 equations in 2 unknowns, v1 and v2. Plug in the values of the known mases, find the initial velocity v0 and plug itin, and solve the equatiolns for v1 and v2.

Get v0 from the formula for velocity of body that falls a distance h:

v = (2 g h )^1/2, (Eqn 3)
where g = 9.80 m/s^2, and h = 1.35 m. So v0 = 5.144 m/s.

The answer to a) is v0 = 5.144 m/s

Next find height of bounce of ball 2 (tennis ball). For this you need the after-collision velocity of ball 2. Get this from equations 1 and 2. Rearrange Eqn 1 to get

v1 = [ v0 (m1 - m2) - m2v2 ] / m1 (Eqn 4)

Plug Eqn 4 into Eqn 2, rearrange to solve for v2.

v1 = [5.144(637-57) - 57v2]/637

5.144^2(637+57) = (2983.52-57v2)^2 +57 (v2)^2

18363.75 = 8901391.59 + 3249 v2^2 - 340121.28v2 + 57v2^2

3306v2^2 - 340121.28v2 + 8883027.84 = 0

v2 = 51.44 m/s

All the kinetic energy of ball 2 will be converted to gravitational potential energy as the ball rises to height h2. Rearrange Eqn 3 to get

h2 = v2^2 / 2g (Eqn 5)

h2 = 51.44^2/2*9.8 = 135 m

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