Question

If the momentum of a body is increased by 50 %, the percentage increase in its kinetic energy is:

• A. 100
• B. 50
• C. 125
• D. 200

Answer 1

The correct option is C 125

Momentum of body : $P=m\times v$
Kinetic Energy of body : $KE=\frac{1}{2}\times m\times v^2$
where, $m=$ Mass of Body
$v=$ Velocity of body
According to question:
Old $P=m\times v$
New $1.5P=m\times v_1$
where $v_1=$new velocity
Hence the value of $v_1=1.5\times v$
$K{E}_{Old}=\frac{1}{2}\times m\times v^2$
$K{E}_{New}=\frac{1}{2}\times m\times {v_1}^2$

Putting in the value of $v_1=1.5\times v$
$K{E}_{New}=1.125\times m\times v^2$
$\Delta KE=0.625\times m\times v^2$
$\frac{\Delta KE}{KE}=1.25\times m\times v^2$
$\Delta KE$% $=125$%