Question

Two person each of mass $m$$arestandingatthetwoextremesofarailroadcarofmassleft$M$restingonasmoothtrack.Thepersonontheleftjumpstowithavelocity$u$. Find the velocity of the train when both the person have jumped off.

• A. $\frac{m^{2}u}{m(m+M)}$ towards left
• B. $\frac{m^{2}u}{M(m+M)}$ towards left
• C. $\frac{m^{2}u}{M(m+M)}$ towards right
• D. $\frac{m^{2}u}{m(m+M)}$ towards right

Answer 1

Answer: (B)

$\frac{m^{2}u}{M(m+M)}$ towards left

From the conservation of linear momentum,

$mu=-(M+m)v$ or $v=\frac{-mu}{M+m}$

Negative sign means towards right

After the 2nd person jumps, conserving momentum (Now taking positive sign from right)

$mu-Mv-M{v}^{\prime }$ or $mu=\frac{M\left(mu\right)}{m+M}-M{v}^{\prime }$

or $v^{\prime }=-\frac{m^{2}u}{M(m+M)}$

Negative sign shows towards left.