Question

A body of mass 60kg has the momentum $3000kgm{{s}^{-1}}$. Calculate the kinetic energy.
A. $7.5\times {{10}^{1}}J$
B. $7.5\times {{10}^{2}}J$
C. $5\times {{10}^{4}}J$
D. $7.5\times {{10}^{4}}J$

Answer 1

Hint: As a very first step, we could note down all the given values from the question. Then we could recall the expression connecting the momentum and kinetic energy of the body. After that all we have to do is simply substitute the given values into this expression and hence find the answer.
Formula used:
Relation between kinetic energy and momentum,
$K.E=\dfrac{{{p}^{2}}}{2m}$

Complete answer:
In the question, we are given a body of mass 60kg and it is said to have a linear momentum of $3000kgm{{s}^{-1}}$ and we are supposed to find the kinetic energy.
Let us recall the relation between kinetic energy of a body with its linear momentum. This expression would be given by,
$K.E=\dfrac{{{p}^{2}}}{2m}$
$\Rightarrow K.E=\dfrac{{{\left( 3000 \right)}^{2}}}{2\times 60}=75000J$
$\therefore K.E=7.5\times {{10}^{4}}J$
Therefore, we found the kinetic energy of the given body to be$K.E=7.5\times {{10}^{4}}J$.

So, the correct answer is “Option D”.

Note: In the above solution we have used the relation that directly finds the kinetic energy from the value of momentum. There is actually an alternate method where we could first find the velocity of the body from the given mass and momentum.
$p=mv=60\times v=3000kgm{{s}^{-1}}$
$\Rightarrow v=\dfrac{3000}{60}=50m{{s}^{-1}}$
Now, we could find the kinetic energy using this velocity that we found as,
$K.E=\dfrac{1}{2}\times 60\times {{\left( 50 \right)}^{2}}$
$\therefore K.E=7.5\times {{10}^{4}}J$
We have gotten exactly the same value that we found at the end of the solution. In this way we could also get to know the value of the velocity of the body.