Question

A particle of mass $m$ moving with kinetic energy $K$, makes a head-on elastic collision with a stationary particle of mass $\eta m$. The maximum potential energy stored in the system during the collision is

• A. $\frac{\eta }{\eta +1}K$
• B. $(\eta -1)K$
• C. $(\eta -1)K$
• D. $\frac{\eta +1}{\eta }K$

Answer 1

The correct option is A $\frac{\eta }{\eta +1}K$

Step 1: Draw a labelled diagram.

Step 2: Find the velocity when particles compress to the maximum.

Formula used:
$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$

During the impact, some part of incident kinetic energy is temporarily stored in the form of potential energy.

The potential energy will be maximum when particles compress to the maximum; at the maximum compression, both particles move with the same velocity.

Applying momentum conservation,

$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$

$mu=(m+\eta m)v\Rightarrow v=\left(\frac{u}{1+\eta }\right)$

Step 3: Find the maximum potential energy stored in the system.

$K_i=\frac{1}{2}m{u}^2$

K.E. at the maximum compression,

$\frac{1}{2}(m+\eta m)v^2=\frac{1}{2}m{u}^2\left(\frac{1}{1+\eta }\right)$

$(P.E.{)}_{max}=\frac{1}{2}m{u}^2-\frac{1}{2}m{u}^2\left(\frac{1}{1+\eta }\right)$

$=\left(\frac{\eta }{\eta +1}\right)\frac{1}{2}m{u}^2=\left(\frac{\eta }{\eta +1}\right)K$