Question

How much work is needed to be done on a ball of mass $50g$ to give it a momentum of $500gcm{s^{ - 1}}$ ?
(A) $10J$
(B) $25 \times {10^{ - 5}}J$
(C) $5 \times {10^4}J$
(D) $0J$

Answer 1

Hint: Here, the ball is needed to be given $500gcm{s^{ - 1}}$ amount of momentum and you are asked to find the work that would be needed to do so. In order to solve this question, consider the definition of work, how it is defined. Also use the relation between the force and momentum to find the work done.

Complete step by step answer:
Consider Newton's second law of motion. It states that the force on an object is equal to mass times the acceleration. Mathematically, $F = ma$. Now, acceleration is defined as the rate of change of velocity with respect to time. So, $F = m\dfrac{{dv}}{{dt}}$. Take the mass inside the derivative as mass is constant, $F = \dfrac{{d\left( {mv} \right)}}{{dt}}$. Product of mass and velocity is defined as momentum and therefore, force is equal to rate of change of momentum with respect to time $ \to F = \dfrac{{dp}}{{dt}}$.

Now, work is given as force dot displacement, mathematically,
$dW = \overrightarrow F .d\overrightarrow r $. Substituting force in the above equation, we get,
$
dW = \dfrac{{d\overrightarrow p }}{{dt}}.d\overrightarrow r \\
\Rightarrow dW = d\overrightarrow p .\dfrac{{d\overrightarrow r }}{{dt}} \\
\Rightarrow dW = d\overrightarrow p .\overrightarrow v \\
\Rightarrow \overrightarrow p = m\overrightarrow v
\Rightarrow\overrightarrow v = \dfrac{{\overrightarrow p }}{m} \\
\Rightarrow dW = d\overrightarrow p .\dfrac{{\overrightarrow p }}{m} \\
\Rightarrow dW = \dfrac{1}{m}pdp \\ $
Let us integrate both sides,
$
\int {dW} = \int {\dfrac{1}{m}pdp} \\
\Rightarrow W = \dfrac{{{p^2}}}{{2m}} \\ $
Given: $m = 50g = 50 \times {10^{ - 3}}kg = 5 \times {10^{ - 2}}$, $p = 500gcm{s^{ - 1}} = 500 \times {10^{ - 3}} \times {10^{ - 2}}kgm{s^{ - 1}} = 5 \times {10^{ - 3}}kgm{s^{ - 1}}$
Therefore,
$W = \dfrac{{{{\left( {5 \times {{10}^{ - 3}}} \right)}^2}}}{{\left( 2 \right)\left( {5 \times {{10}^{ - 2}}} \right)}} \\
\therefore W = 25 \times {10^{ - 5}}J$
Hence, work needed to be done on a ball of mass $50g$ to give it a momentum of $500gcm{s^{ - 1}}$ will be $25 \times {10^{ - 5}}J$.

Hence, option B is correct.

Note: The point to be noted is that we have not taken the integration constant into account. Actually, we have considered it, but as the initial momentum is to be zero and so will be the work done initially, the integration constant will come out to be zero. Also keep in mind the various relations we have used as, relation between force and work, relation between force and momentum and relation between acceleration and velocity.