Question 257: Centre of Mass and Rotational Motion

Question

The momentum of a body is increased by $$50 \%$$. The percentage increase in the kinetic energy of the body is ___________ $$\%$$.

Accepted Answer

The momentum (p) and kinetic energy (K) of a body are related by the equations:
$p = mv$,
$K = \frac{1}{2}mv^2$,
where m is the mass and v is the velocity of the body.
We can express v in terms of p and m:
$v = \frac{p}{m}$,
and substitute this into the equation for K to get:
$K = \frac{p^2}{2m}$.
So, the kinetic energy is proportional to the square of the momentum.
If the momentum is increased by 50%, the new momentum is 1.5p, and the new kinetic energy is:
$K' = \frac{(1.5p)^2}{2m} = \frac{2.25p^2}{2m} = 2.25K$.
The percentage increase in the kinetic energy is then:
$\frac{K'-K}{K} 	imes 100 = \frac{2.25K - K}{K} 	imes 100 = 1.25 	imes 100 = 125\%$.
So, the percentage increase in the kinetic energy of the body is 125%.

Centre of Mass and Rotational Motion

The momentum of a body is increased by $50 \%. The percentage increase in the kinetic energy of the body is ___________ \%$.

Integer / Numeric Answer Type