Question

A ball of mass m is moving with a speed v as shown in the figure. It undergoes inelastic collision with a ball of mass $2\text{m}$ which was initially at rest. The velocity of ball $2\text{m}$ after collision is $\frac{\sqrt{\text{p}}v}{\text{q}}$. Find $\text{p}+\text{q}$.

Answer 1

Draw a labelled diagram.


Find the of velocity along the line of impact.
From the diagram,
$\sin \theta =\frac{\text{R}}{2\text{R}}=\frac{1}{2}={30}^{\circ }$
Find the velocity after collision.
From the law of conservation of momentum,
${\text{m}}_1{\text{u}}_1+{\text{m}}_2{\text{u}}_2={\text{m}}_1{\text{v}}_1+{\text{m}}_2{\text{v}}_2$
$\text{mv}\cos {30}^0={\text{mv}}_2+2{\text{mv}}_1...\left(i\right)$
Apply Coefficient of Restitution.
$\text{e}=\frac{1}{2}$
$\text{e}=\frac{1}{2}=\frac{{\text{v}}_1-{\text{v}}_2}{\text{v}\cos {30}^0}..\left(ii\right)$
On solving (i) and (ii),
${\text{v}}_1=\frac{\sqrt{3}v}{4}$
Final Answer: 7