Question

A ball A of mass m falls under gravity from a height h and strikes another ball B of mass m which is supported at rest on a spring of stiffness k. Assume perfect elastic impact. Immediately after the impact

• A. The velocity of ball A is $\frac{1}{2}\sqrt{2gh}$
• B. The velocity of ball A is Zero
• C. The velocity of both balls is $\frac{1}{2}\sqrt{2gh}$
• D. None of the above

Answer 1

The correct option is B The velocity of ball A is Zero

Method I:

In a perfectly elastic collision velocity of the ball a will be zero after impact. Because in perfectly elastic impact velocities are interchanged when masses are the same.
Method II:
$e=1\to$ in a perfectly elastic collision.


Before impact velocity of ball A

$u_A=\sqrt{2gh}\text{and}{u}_B=0$

$\text{Vel. of app.}=u_A-u_B$
$=\sqrt{2gh}$

Conserving linear momentum
$m_A{u}_A+0=m_A{v}_A+m_B{v}_B$

$m_A=m_B=m$

$u_A=V_A=V_B...\left(i\right)$

For elastic collision velocity of approach = velocity of separation

$u_A-u_B=V_B-V_A...\left(ii\right)$

Adding Eqs. (i) and (ii), we get

$2u_A-u_B=2V_B$

As $u_B=0$

$\Rightarrow 2u_A=2V_B$

$u_A=V_B$

Using value of $u_A$ in Eq. (i), we get

$V_A=0$
$V_B=u_A=\sqrt{2gh}$