Question

Two men jump from a stationary buggy, one after the other with the same velocity $u=20\text{m/s}$ relative to the buggy towards left. Masses of the men are same ($m=50\text{kg}$) & mass of the buggy is $M=200\text{kg}$. Find the velocity of the buggy after both men jump off. (Assume friction to be negligible )

• A. $\frac{70}{3}\text{m/s}$ towards left
• B. $\frac{110}{9}\text{m/s}$ towards left
• C. $\frac{22}{3}\text{m/s}$ towards right
• D. $\frac{220}{9}\text{m/s}$ towards right

Answer 1

The correct option is C $\frac{22}{3}\text{m/s}$ towards right

Given,
Mass of buggy $=M=200\text{kg}$
Mass of each man $=m=50\text{kg}$
and relative speed of jumping $u=20\text{m/s}$ (w.r.t buggy)

As no external force is acting, so momentum will be conserved.
i.e $P_i=P_f$


Here $v$ = velocity of buggy after first man has jumped. Taking leftward direction as +ve,
$P_i=P_f$
$\Rightarrow 0=m(u-v)-(M+m)v$
$\Rightarrow mu-mv=(M+m)v$
$\Rightarrow v=\left(\frac{mu}{M+2m}\right)=\left(\frac{50\times 20}{200+2\times 50}\right)=\frac{10}{3}\text{m/s}$

After second man jumps:


Here, $v_1$ = velocity of buggy after second man jumps
Taking leftward direction as +ve,
$P_i=P_f$
$\Rightarrow -(M+m)v=m(u-v_1)-M{v}_1$
$\Rightarrow -(M+m)v=mu-m{v}_1-M{v}_1$
$\Rightarrow -(m+M)v-mu=-(m+M)v_1$
$\Rightarrow -(50+200)\times \frac{10}{3}-50\times 20=-(50+200)v_1$
$\Rightarrow (50+200)(v_1-\frac{10}{3})=1000$
$\Rightarrow v_1=4+\frac{10}{3}=\frac{22}{3}\text{m/s}$
$\therefore v_1=\frac{22}{3}\text{m/s}$ (towards right) is the final velocity of buggy.