Question

A cart of mass $M$ has a pole on it from which a ball of mass $\mu$ hangs from a thin string attached to point P. The cart and the ball have initial velocity $v$. The cart crashes onto another cart of mass $m$ and sticks to it. The length of the string is $R$. If the smallest initial velocity from which the ball can go in the circle around point P is $v$, find the value of $v$. (Neglect friction and take $M$, $m>>$ $\mu$)

• A. $\left(\frac{M+m}{M}\right)\sqrt{4gR}$
• B. $\left(\frac{M+m}{m}\right)\sqrt{4gR}$
• C. $\left(\frac{M+m}{m}\right)\sqrt{5gR}$
• D. $\left(\frac{M+m}{M}\right)\sqrt{5gR}$

Answer 1

Answer: (C)

$\left(\frac{M+m}{m}\right)\sqrt{5gR}$

$\mu <<$m, M

Momentum conservation gives: $Mv=(M+m)v^{\prime }$
$\Rightarrow$ velocity of the two carts after collision $v^{\prime }=\frac{Mv}{M+m}$
Consider the circular motion of the ball atop the cart if it were stationery. If at the lowest and highest points the ball has speeds $v_1$ and $v_2$, respectively, then
$\frac{1}{2}\mu v_1^2=\frac{1}{2}\mu v_2^2+2\mu gR$
and $\frac{\mu v_2^2}{R}=T+\mu g$,
where $T$ is the tension in the string when the ball is at the highest point.
The smallest $v_2$ is given by $T=0$.
$\Rightarrow$ The smallest $v_1$ is given by
$\frac{1}{2}\mu v_1^2=\frac{1}{2}\mu gR+2\mu gR$
$\Rightarrow v_1=\sqrt{5gR}$
When the cart is moving, $v_1$ is the velocity of the ball relative to the cart. As the ball has initial velocity, velocity of the ball relative to the cart after the collision is $(v-v^{\prime })$. Hence, the smallest velocity for the ball to go round in a circle after the collision is given by:
$v_1=(v-v^{\prime })=v-\left(\frac{Mv}{M+m}\right)=\sqrt{5gR}$
$\Rightarrow v=\left(\frac{M+m}{m}\right)\sqrt{5gR}$