Question

Two balls having masses m and 2m are fastened to two light strings of same length l (figure 9-E18). The other ends of the strings are fixed at O. the strings are kept in the same horizontal line and the system is released from rest. The collision between the balls is elastic. (a) Find the velocities of the balls just after their collision. (b) How high will the balls rise after the collision ?

Answer 1

Let the velocity of m reaching at lower end = $v_1$

From work energy principle.

$\therefore \left(\frac{1}{2}\right)\times m{v}_1^2-\left(\frac{1}{2}\right)\times m(0{)}^2=mgl$

$\Rightarrow v_1=\sqrt{2gl}$

Similarly velocity of heavy block will be,

$v_2=\sqrt{2gl}$

$\therefore v_1=v_2=u$ (say)

Let final velocity of m and 2m are $v_1$ and $v_2$ respectively

According to law of conservation of momentum,

$\Rightarrow m\times u-2mu=m{v}_1+2m{v}_2$

$\Rightarrow v_1+2v_2=-u$

Again $v_1-v_2=[u-v]=-2u$

Substracting,

$3v_2=u$

$\Rightarrow v_2=\frac{u}{3}=\sqrt{\frac{2gl}{3}}$

Substituting in (ii)

$v_1-v_2=-2u$

$\Rightarrow v_1=-2u+v_2$

$=-2u+\left(\frac{u}{3}\right)$

$=\frac{-5}{3}u=\frac{-5}{3}\times \sqrt{2gl}$

$=\frac{\sqrt{50gl}}{3}$

(b) Putting the work energy principle

$\left(\frac{1}{2}\right)\times 2m\times (0{)}^2-\left(\frac{1}{2}\right)\times 2m\times (v_2{)}^2=-2m\times g\times h$

$[h\to$ height reached by heavy ball]

$\Rightarrow h=\left(\frac{L}{9}\right)$

Similarly

$\left(\frac{1}{2}\right)\times m\times (0{)}^2-\left(\frac{1}{2}\right)\times m{v}_1^2=m\times g\times h_2$

[height reached by small ball]

$=\left(\frac{1}{2}\right)\times \frac{50gL}{9}g\times h_2$

$h_2=\frac{0.25L}{9}$

Some $h_2$ is more than 2L, the velocity at highest point will not be zero.

And the 'm' will rise by a distance 2L.