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 a box. Find out the velocity of the box after the person jumps on?
(A) $12.5 m/s$
(B) $25.0 m/s$
(C) $35.7 m/s$
(D) $83.3 m/s$

Answer 1

Hint
These types of questions can be solved using the formulas. The equations which established the relation between initial velocity, final velocity, time, the acceleration of the body and displacement are known as the kinematic equation of motions.
Using, $v = u + at$
$u = $initial velocity $(m/s)$
$v = $final velocity $(m/s)$
$a = $acceleration of the body $(\mathop {ms}\nolimits^{ - 2} )$
$t = $the time is taken $(s)$

Complete step by step solution
According to question,
Given: \[mass\left( M \right) = 70.0{\text{ }}kg\]
\[Force\left( F \right) = {\text{ }}500{\text{ }}N\]
\[m = {\text{ }}30.0{\text{ }}kg\]
\[t{\text{ }} = {\text{ }}5{\text{ }}sec\]
Initial velocity of the box, $u = 0m/s$
Acceleration of the box,
$a = \dfrac{F}{m} = \dfrac{{500}}{{30}} = \dfrac{{50}}{3}m/s$
Using, $v = u + at$
$u = $initial velocity $(m/s)$
$v = $final velocity $(m/s)$
$a = $acceleration of the body $(\mathop {ms}\nolimits^{ - 2} )$
$t = $the time is taken $(s)$
$u = 0m/s$
$a = \dfrac{{50}}{3}m/s$
Then, $v = 0 + \dfrac{{50}}{3} \times 5$
$ \Rightarrow v = \dfrac{{250}}{3}m/s$
So, the total momentum of the system just before the jump,
$\mathop P\nolimits_i = (M + m)v$
$\mathop P\nolimits_i = $Initial momentum
$\mathop P\nolimits_i = (70 + 30) \times \dfrac{{250}}{3} = 2500kgm/s$
Let the velocity of the ‘man+box’ system after the jump be $V$
Then the final momentum after the jump will be=
$\mathop P\nolimits_f = (M + m)V$
$\mathop P\nolimits_f = $Final momentum
$\mathop P\nolimits_f = (M + m)V = 100V$
( where \[V\] is the common velocity with which they both will move after jump)
Now, we are using the momentum conservation,
$\mathop P\nolimits_i = \mathop P\nolimits_f $
Initial momentum = final momentum
Therefore, $2500 = 100V$
$ \therefore V = \dfrac{{2500}}{{100}} = 25m/s$
So, the velocity of the box after the jumps on will be $25m/s$.
So, the correct answer is option (B).

Note
For these types of questions, we need to have a clear understanding of all the three equations of motions. We need to be clear with the concept of forces, and conservation of momentum, and how to calculate them. The law of conservation of momentum states that total momentum before collision and after collision remains conserved. It can be transformed to one body to another but for the whole system it remains conserved before and after Collision.