Question

KE of a body increased by 44%. What is the percentage increase in the momentum?

• A. 10%
• B. 20%
• C. 30%
• D. 44%

Answer 1

The correct option is B 20%

Kinetic Energy, $K.E=\frac{1}{2}m{v}^2$

Momentum, $P=mv$

$K.E=\frac{P^2}{2m}......\left(1\right)$

After increasing 44 % in kinetic energy.

$(1+\frac{44}{100})K.E=\frac{P_{new}^2}{2m}....\left(2\right)$

$\frac{P_{new}}{P}=\sqrt{\frac{1.44}{1}}=1.2$

$P_{new}=1.2P=(1+\frac{20}{100})P$

$P_{new}-P=\frac{20}{100}P$

Hence, momentum increase 20 %