Question

The loss in kinetic energy of the system during collision is

• A. $\frac{7m{v}_0^2}{27}$
• B. $\frac{15m{v}_0^2}{64}$
• C. $\frac{12m{v}_0^2}{25}$
• D. none of these

Answer 1

The correct option is A $\frac{7m{v}_0^2}{27}$

After collision of C with B , B and C will move together

$m{v}_o=(m+m)v$

$v=\frac{v_o}{2}$

Now when the string will get taut , then the velocity of A and B,C will be same along the cord ,

let $u=$ velocity along the cord of A, B and C

So $2mvsin\theta =(m_A+m_B+m_c)u$

So $u=\frac{v_{o}sin\theta }{3}$ , here $\sin \theta =\frac{l/3}{l}=\frac{1}{3}$

So $u=\frac{v_o}{9}$

that is the velocity of A

$u$ is the component of velcoity of B and C along cord.

The velocity perpendicular to cord will remain same

So $v_{per}=vcos\theta$ , $cos\theta =\frac{2\sqrt{2}}{3}$

$v_{per}=\frac{v_o\sqrt{2}}{3}$

total velocity of B and C

$V^2=u^2+v_{per}^2=$

$V=\frac{v_o\sqrt{19}}{9}$

So Initial $K{E}_i$ =$\frac{m{v}_o^2}{2}$

final $K{E}_f=\frac{m{u}^2}{2}+\frac{2m{V}^2}{2}=\frac{39m{v}_o^2}{162}$

So loss in KE is $K{E}_i-K{E}_f=\frac{7m{v}_o^2}{27}$