Question

Two balls of masses $5\text{kg}$ and $10\text{kg}$ are at the positions shown in figure. The track on which the balls move is frictionless. Initially, the $10\text{kg}$ ball is kept at rest and the $5\text{kg}$ ball is dropped with speed $v$ from height $5\text{m}$. Assuming the collision between the balls is perfectly elastic, find the value of ${}^{\prime }{v}^{\prime }$ such that the $10\text{kg}$ ball reaches point ${}^{\prime }{C}^{\prime }$.

• A. $10\text{m/s}$
• B. $0\text{m/s}$
• C. $22.4\text{m/s}$
• D. $11.2\text{m/s}$

Answer 1

The correct option is D $11.2\text{m/s}$

Since the track is frictionless, we can apply the law of conservation of energy between $A$ and $B$
For ball of mass $m$:
$K_A+U_A=K_B+U_B$
$\frac{1}{2}m{v}^2+mg\left(h\right)=\frac{1}{2}m{u}_1^2+0$
(taking $U_B=0$)
$\Rightarrow u_1^2=v^2+2\left(10\right)\left(5\right)=v^2+100...\left(1\right)$

[If $5\text{kg}=m$, then $10\text{kg}=2m$]

Applying law of conservation energy between $B$ and $C$
For ball of mass $2m$:
$K_B+U_B=K_C+U_C$
$\frac{1}{2}\left(2m\right)v_2^2+0=\left(2m\right)gh+0$
[for minimum value of $v$, the ball just reaches $C$ i.e $K_C=0$]
$\Rightarrow v_2^2=2gh=100....\left(2\right)$

Given that the collision is perfectly elastic, velocity of ball of mass $2m$ (i.e $m_2$) is given by
$v_2=\left(\frac{m_2-m_1}{m_1+m_2}\right)v_1+\left(\frac{2m_1}{m_1+m_2}\right)u_1$
$=\frac{2m}{m+2m}\times u_1$
($\because$ initial velocity of ball $m_1=u_1$
& initial velocity of ball $m_2,v_1=0$)
$\Rightarrow v_2=\frac{2u_1}{3}......\left(3\right)$

From (3) and (2)
$\left(\frac{4u_1^2}{9}\right)=100$
$\Rightarrow u_1^2=\frac{900}{4}=225.......\left(4\right)$

From $\left(1\right)$ and $\left(4\right)$
$225=v^2+100$
$\Rightarrow v^2=125$
$\Rightarrow v=11.2\text{m/s}$ is the velocity with which the ball should be dropped.