Question

A man wearing a bulletproof vest stands still on roller skates. The total mass is $80{\text{ }}kg$ . A bullet of mass $20{\text{ }}grams$ is fired at $400\,m{s^{ - 1}}$ . It is stopped by the vest and falls to the ground. What is the velocity of the man?

Answer 1

To answer the above issue regarding the velocity of the man, we will first place all of the given data in the correct order, and then utilise the principle of conservation of momentum to solve and finalise our response, that is, initial momentum is equals final momentum.

Complete step by step answer:
Now that we've been provided some information in response to the query, let's take a look at it and write it down. Here, mass of the bullet $= 20\,g$. As the mass is given in grams we will convert it into kilograms.
$\text{mass of the bullet (m)}= \dfrac{{20}}{{1000}}kg \\\ \Rightarrow \text{mass of the bullet (m)}= 0.02\,kg$

Now, the initial velocity of the bullet is given as $400\,m{s^{ - 1}}$.
Momentum of the bullet before hitting the man is $= m \times u = 0.02 \times 400 = 8\,kg \cdot m{s^{ - 1}}$
(Where, $m$ stands for the mass of the bullet and $u$ stands for the initial velocity of the bullet)
The bullet grinds to a stop after striking the man and falls to the ground.And now, mass of the man is given as $80kg$.If $v$ is the velocity of the man after being hit by the bullet then,
$M \times v = 8 \\\ \therefore v = \dfrac{8}{{80}} = 0.1\,m{s^{ - 1}}$

Therefore, the required velocity of the man is $0.1\,m{s^{ - 1}}$.

Note: Students should take care of the conditions while doing the numerical based on conservation of momentum that, when the mass of the system in the given question remains constant during the contact and no net external force occurs on the system during the interaction, momentum is conserved.