Question

You have two balls of identical mass and size that behave very differently. When you drop the so-called “sad” ball, it thuds on the floor and does not bounce at all. When you drop the so-called “happy” ball from the same height, it bounces back to almost the same height from which it was dropped. The difference in the bouncing ability of the happy ball is due its internal structure; it is made of different material. You hang each ball from a string of identical length and place a wood board on its end directly below the support for each string(shown below). You pull each ball back to an equal height and release the balls one at a time. When each ball hits the board, which has the best chance of knocking the board over: the sad ball or the happy ball? 

Answer 1

Explanations & Calculations

• According to the description, the coefficient of restitution ($\small e$ ) between the happy ball & the floor can be thought to equal to almost 1 ($\small e= 1$ ) as the ball bounces back almost to the same height.
• The collision then appears to be elastic where kinetic energy after the collision equals that before the collision.
• If we assume the happy ball was approaching the floor at some velocity $\small \downarrow v$, then at which it would bounce back will be $\small \uparrow ev$. (in this case, it is almost $\small \uparrow v$ )

• Accordingly, the collision of the sad ball with the floor appears to be completely inelastic where $\small e=0$.

• It is apparent that there involves a change in the linear momentum of both balls during the collisions.
• Change in momentum results in generating an impulse on the balls & equally & oppositely on the floor(where they collide).
• Impulse is a large force acting in a short period of time thus able to perform some work.

• Let's assume that the balls approach the wood board with some velocity $\small v_1$
• Note that $\small e$ depends on the nature of the surfaces, hence let's take that between the balls & the wood board to be $\small e_1$

• The impulse generated from the happy ball is

$\qquad\qquad \begin{aligned} \small I_h &= \small \Delta mv=m\Delta v\\ \small I_h &= \small m[\uparrow e_1v_1-(-\downarrow v_1)]=m[\uparrow e_1v_1+\uparrow v_1]\\ &=\small \bold{\uparrow mv(e+1)}\cdots(on\,the\,board\,is\,\downarrow mv(e+1)) \end{aligned}$

• From the sad ball,

$\qquad\qquad \begin{aligned} \small I_s &=\small m[0-(\downarrow v_1)]\cdots(e_1\approx0\implies e_1v_1\to 0)\\ &= \small \bold{\uparrow mv_1}\cdots(on\,the\,board\,is\,\downarrow mv_1) \end{aligned}$

• It is obvious that $\small I_{happy}>I_{sad}$

• Therefore, flipping the board over would be possible with the happy ball.