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

Question

Accepted Answer

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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:

$\frac{K'-K}{K} \times 100 = \frac{2.25K - K}{K} \times 100 = 1.25 \times 100 = 125\%$.

So, the percentage increase in the kinetic energy of the body is 125%.

"

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