Question

A small ball of mass $m$ is aligned above a larger ball of mass $M = 0.63 kg$ (with a slight separation, as with the baseball and basketball in figure shown below), and the two are dropped simultaneously from a height of $h = 1.8 m$ . (Assume the radius of each ball is negligible relative to $h$ ) (a) If the larger ball rebounds elastically from the floor and then the small ball rebounds elastically from the larger ball, what value of $m$ results in the larger ball stopping when it collides with the small ball?
(b) What height does the small ball then reach in the figure shown below: [ take $g = 10 m / s^{2}$ ]

Accepted Answer

Given, Mass of small ball $= m$
Mass of large ball $= M = 0.63 kg$

Let's denote,

Velocity of both the balls after falling through height $h$ , just before striking the floor $h$
After collision between the balls, velocity of small ball $= v_{1}$
Velocity of large ball $= v_{2} = 0$

After falling through height $h$ , velocity gained by each ball, $v = \sqrt{2 g h} = \sqrt{2 \times 10 \times 1.8} = 6 m s^{- 1}$

(Using energy conservation : $\frac{1}{2} m v^{2} = m g h$ )

Large ball collides with the floor elastically and bounces back with velocity $v$ upward, small ball is moving with velocity $v$ downward and going to collide with the large ball elastically. The situation is as shown in the figure below:

After elastic collision between the balls, small ball moves upwards with velocity $v_{1}$ and the large ball comes to momentary rest ( $v_{2} = 0$ ). The situation is as shown in the figure below:

(a)For elastic collision between the balls:

Applying momentum conservation, $M v - m v = m v_{1} . . . (1)$

As the collision is elastic approach and separation speeds can be equated, $v_{1} = v - (- v) = 2 v = 12 m s^{- 1}$ $. . . (2)$

Substituting, $v = 6 m s^{- 1}$ , $v_{1} = 2 v = 12 m s^{- 1}$ and $M = 0.63 kg$ in equation $(1)$
$\Rightarrow m = 0.21 kg$

(b) Let $h_{1}$ be the maximum height attained by the small ball

Applying energy conservation between the floor level and maximum height level:

$m g h_{1} = \frac{1}{2} m {v_{1}}^{2}$

$\Rightarrow h_{1} = \frac{{v_{1}}^{2}}{2 g} = \frac{{(12)}^{2}}{2 \times 10} = 7.2 m$

Resnick & Halliday Solutions for Chapter: Center of Mass and Linear Momentum, Exercise 1: Problems

Resnick & Halliday Physics Solutions for Exercise - Resnick & Halliday Solutions for Chapter: Center of Mass and Linear Momentum, Exercise 1: Problems

Questions from Resnick & Halliday Solutions for Chapter: Center of Mass and Linear Momentum, Exercise 1: Problems with Hints & Solutions

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