Question

If the momentum of a body is increased by 50%, what
will be the percentage increase in the kinetic
energy of the body?​

Answer 1

Therefore, the percentage increase in its kinetic energy is 125%

hope it helps

Answer 2

Answer:

The percentage increase in kinetic energy of the body is 125%

Explanation:

We must know the relation between Kinetic Energy and momentum to find the percentage increase in kinetic energy of the given body.

Kinetic Energy is given by,

 $\boxed{\bf KE=\dfrac{1}{2}mv^2}$

momentum of a body is the product of mass and velocity.

P = mv

So,

 $\sf KE=\dfrac{1}{2}mv \times v \\\\ \sf KE=\dfrac{1}{2}P \times \dfrac{v \times m}{m} \\\\ \sf KE=\dfrac{1}{2} \times \dfrac{P^2}{m} \\\\ \boxed{\sf KE=\dfrac{P^2}{2m}}$

⇒ KE ∝ P²

$\longrightarrow \bf \dfrac{KE_2}{KE_1}=\dfrac{P_2^2}{P_1^2} \\\\ \longrightarrow \sf \dfrac{KE_2}{KE_1}-1=\dfrac{P_2^2}{P_1^2}-1 \\\\ \longrightarrow \sf \dfrac{KE_2-KE_1}{KE_1}=\dfrac{P_2^2}{P_1^2} -1$

% increase in Kinetic Energy of the body is :

  $\sf \% \ increase \ in \ KE=\bigg( \dfrac{P_2^2}{P_1^2} -1 \bigg) \times 100$

Let the initial momentum of the body be, P₁ = 100

when the momentum is increased by 50% , the final momentum is

P₂ = 100 + 50% of 100

P₂ = 100 + 50

P₂ = 150

% increase in Kinetic Energy of the body :

 $\sf =\bigg( \dfrac{150^2}{100^2}-1 \bigg) \times 100 \\\\ \sf =\bigg( \dfrac{22500}{10000}-1 \bigg) \times 100 \\\\ \sf =\bigg( \dfrac{225}{100}-1 \bigg) \times 100 \\\\ \sf =(2.25-1) \times 100 \\\\ \sf =1.25 \times 100 \\\\ \sf =125\%$

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