Question

Two steel balls of equal diameter are connected by a rigid bar of negligible weight as shown and are dropped in the horizontal position from height $h$ above the heavy steel and brass base plates. If the coefficient of restitution between the ball and the steel base is $0.6$ and that between the other ball and the brass base is $0.4$, the angular velocity of the bar immediately after rebound is (Assume the two impacts are simultaneous and take $g=10{\text{m/s}}^2$).

• A. $\frac{2}{5}\text{rad/s}$
• B. $\frac{1}{5}\text{rad/s}$
• C. $\frac{3}{5}\text{rad/s}$
• D. $\frac{1}{4}\text{rad/s}$

Answer 1

The correct option is A $\frac{2}{5}\text{rad/s}$

If $v$ is the velocity with which the balls strike the bases, then their velocities after collision are $0.6v$ and $0.4v$ as shown.
Let ${}^{\prime }{u}^{\prime }$ be the velocity of $COM$ just after collision and $\omega$ is the angular velocity acquired by the rod. Since diameters of the steel balls are same, their masses are also same and hence $COM$ will be at $\frac{l}{2}$ from $A$


For end A: $0.6v=u+\frac{1}{2}\cdot \omega l$ ...(i)
For end B: $0.4v=u-\frac{1}{2}\cdot \omega l$ ...(ii)
Subtracting (i) and (ii) we get
$0.2v=\omega l\Rightarrow \omega =\frac{v}{5l}$

Now using $[v^2=u^2+2as]$
Velocity of balls before collsion, $v$ is given by
$v^2=0+2\times 10\times 0.2=4$
$\Rightarrow v=2\text{m/s}$
or $\omega =\frac{2}{5\times 1}=\frac{2}{5}\text{rad/sec}$