Question

Two pedulum bobs of mass $m$ and $2m$ collide elastically at the lowest point in their motion. If both the balls are released from a height $H$ above the lowest point, to what heights do they rise for the fist time after collision?

Answer 1

Let they obtain speed v just before collision

$\therefore v=\sqrt{2gH}$

Let them both move in left direction with speed $v_1$ & $v_2$

Momentum conservation $=P_1(2mv-mv$ & $mv)$

$P_f=m{v}_1+2m{v}_2$

$\Rightarrow v=v_1+2v_2$ .........$\left(1\right)$

Energy conservation: $K{E}_{initial}=K{E}_{final}=PE$ before release

$\therefore 3mgH=\frac{1}{2}m{v}_1^2+\frac{1}{2}2m{v}_2^2$

$6gH=v_1^2+2v_2^2$

for case, put $2gh=v^2$

$3v^2=v_1^2+2v_2^2$ ..........$\left(2\right)$

Solving $\left(1\right)$ & $\left(2\right)$ we get

$v=v$

$v_1=-v$

As magnitude of velocitues is same, they rise to same height i.e., H.