Question

A bullet of mass $60\;{\rm{g}}$ moving with a velocity of $500\;{\rm{m}}{{\rm{s}}^{ - 1}}$ is brought to rest in $0.01\;{\rm{s}}$. Find the impulse and the average force of the blow.
(A) $- 3\;{\rm{Ns, 3}}\;{\rm{kN}}$
(B) $- 30\;{\rm{Ns, - 3}}\;{\rm{kN}}$
(C) $- 50\;{\rm{Ns, - 5}}\;{\rm{kN}}$
(D) $- 60\;{\rm{Ns, - 6}}\;{\rm{kN}}$

Answer 1

In this solution, we will apply the impulse momentum theorem.
The impulse equation will give the average force.

Complete step by step answer:
Given,
The initial velocity of the bullet, $u = 500\;{\rm{m}}{{\rm{s}}^{ - 1}}$
The time of impact of the impulse, $t = 0.01\;{\rm{s}}$
The mass of the bullet,
$\begin{array}{c} m = 60\;{\rm{g}}\\\ = {\rm{60}} \times {\rm{1}}{{\rm{0}}^{ - 3}}\;{\rm{kg}} \end{array}$
Here, a bullet moving with a particular velocity comes to rest due to an impulse.

According to the impulse momentum theorem, an impulse acting on an object causes a change in momentum of the object. If $I$ is the impulse acted on the bullet and $p$ is the momentum, we can write
$I = \Delta p$
Here $\Delta p$ is the change in momentum.

Since momentum is the product of the mass and the velocity of an object, we can write the momentum of the bullet as
$p = mv$

Hence,
$I = \Delta \left( {mv} \right)$

Since mass $m$ is a constant,
$I = m\Delta v$ Here $\Delta v$ is the change in velocity.

The change in velocity is,
$\Delta v = v - u$
Here $v$ is the final velocity.

Since the bullet comes to rest, its final velocity $v$ is $0$. As the initial velocity $u$ is $500\;{\rm{m}}{{\rm{s}}^{ - 1}}$, the change in velocity becomes,
$\begin{array}{c} \Delta v = 0 - 500\\\ = - 500\;{\rm{m}}{{\rm{s}}^{ - 1}} \end{array}$

Now, substituting the value of $m$ and $\Delta v$ in $I = m\Delta v$, we get
$\begin{array}{c} I = {\rm{60}} \times {\rm{1}}{{\rm{0}}^{ - 3}}\; \times \left( { - 500\;} \right)\\\ = - 30\;{\rm{Ns}} \end{array}$

Now, another equation for the impulse can be written as
$I = Ft$

Here $F$ is the average force of impact.
Now, substituting the values of $I$ and $t$ in the above equation, we get

$I = Ft\\\ \implies - 30\; = F \times 0.01\;\\\ \implies F = \dfrac{{ - 30\;}}{{0.01\;}}\\\ \implies F = - 3000\;{\rm{N}}\;$
$\implies F = - {\rm{3}}\;{\rm{kN}}$
Therefore, the average force of the blow is $- 3\;{\rm{kN}}$.

Since the values of impulse and average force are $- 30\;{\rm{Ns}}$ and $- 3\;{\rm{kN}}$

So, the correct answer is “Option B.

Note:
Impulse on an object is the impact caused by a force acting on it for a short span of time. In a sense, impulse is also a force. Force exerted on an object is mass times the acceleration of the object. This force law is given by Newton's second law.