Question

A person of mass $70.0kg$ applies a force of $500N$ to a $30.0kg$ box, which is initially at rest on a frictionless surface, over a period of $5.00s$ before jumping onto the box. Find out the velocity of the box after the person jumps on?

• A. $12.5m/s$
• B. $25.0m/s$
• C. $35.7m/s$
• D. $83.3m/s$
• E. $100m/s$

Answer 1

Answer: (B)

$25.0m/s$

Given : $M=70kg$ $F=500N$ $m=30kg$ $t=5s$

Initial velocity of the box $u=0$ m/s

Acceleration of the box $a=\frac{F}{m}=\frac{500}{30}=\frac{50}{3}$ $m/s^2$

Using $v=u+at$

$\therefore$ $v=0+\frac{50}{3}\times 5=\frac{250}{3}$ $m/s$

Total momentum of the system just before the jump $P_i=70\times 0+30\times \frac{250}{3}=2500$ $kg$ $m/s$

Let the velocity of the "man+ box" system after the jump be $V$.

Thus total momentum of the system just after the jump $P_f=(m+M)V=100V$

Using conservation of momentum : $P_i=P_f$

$\therefore$ $2500=100V$ $⟹V=25$ $m/s$