Question

2. Professor Redmount's Last Lecture: A racquetball (diameter 8.00 cm, mass 60.0 g) is held a negligible distance above a basketball (diameter 30.0 cm, mass 610. g). The bottom of the basketball is 4.00 m above a hard floor, e.g., at the height of the second-floor railing in the rotunda of McDonnell-Douglas Hall. The two balls are dropped together from rest. The basketball bounces off the floor and then strikes the racquetball All collisions have 0.700 coefficient of restitution. Neglecting aerodynamic effects, calculate the height to which the racketball would rebound, if the ceiling of the rotunda were not an obstacle.

Answer 1

Since the bottom of basket ball is 4 m high from the ground, both the racquetball and basket ball will drop $H=4\;m$ before rebounding. Also at the time of rebound, the racquetball center would be at certain height $(h_i)$ from ground given by:

Now, velocity attained by the basketball before rebound (down ward direction is negative):

$u_B=-\sqrt{2gH}$

Coefficient of restitution is 0.7 and given by:

$e=0.7=-\frac{v_B}{u_B}\Rightarrow v_B=0.7\sqrt{2gH}$

Now, velocity attained by the racquetball before rebound (down ward direction is negative):

$u_R=-\sqrt{2gH}$

Now since, racquetball hit the basket ball which also has velocity in upward direction, we will have to use the relative velocity for coefficient of restitution.

$e=0.7=-\frac{v_R-v_{BR}}{u_R-u_{BR}}$.......EQ1

where;$u_{BR}=v_B=0.7\sqrt{2gH}$ is the velocity of basketball after rebound but before collision with racquetball. We will keep upward direction as positive.

$v_{BR}$ is the velocity of basketball after collision with racquetball.

rewriting EQ1, we get

$0.7u_R-0.7u_{BR}=v_{BR}-v_R\;\;\Rightarrow v_{BR}=v_R+0.7u_R-0.7u_{BR}$...EQ2

$\Rightarrow v_{BR}=v_R-0.7\sqrt{2gH}-0.49\sqrt{2gH}$.......EQ3

Now conservation of momentum, final minus initial momentum is zero,

$\left (m_Rv_R+m_Bv_{BR} \right )-\left (m_Ru_R+m_Bu_{BR} \right )=0$

using EQ 2

$m_Rv_R+m_B\left [v_R+0.7u_R-0.7u_{BR} \right ] -m_Ru_R-m_Bu_{BR} =0$

substituting values we get

$(60+610)v_R+(0.7\times610 -60)(-\sqrt{2gH})-(1.7\times610 )\times0.7\sqrt{2gH} =0$

$670v_R-367\sqrt{2\times9.81\times4}-725.9\sqrt{2\times9.81\times4} =0$

$v_R=\frac{9681.885}{670}=14.450 \;m/s$

This velocity will convert to a height equal:

$v_R=\sqrt{2gh}\;\;\Rightarrow h=\frac{v_R^2}{2g}=\frac{14.45^2}{2\times9.81}=10.642\; m$

The total height of rebound will be:

$h_{total}=h+h_i=10.642+0.34=\boldsymbol{10.982\;m}$