Question

If the kinetic energy of a body is increased by 300 %, the momentum of the body is increased by

• A. 50%
• B. 100%
• C. 200%
• D. 300%

Answer 1

Answer: (B)

: We know that $p^2=$ link where p denotes momentum and k denotes kinetic energy of the body. Let initial $K.E=100A:$ $\therefore$ Finial $K.E=400A;$ $\therefore$ $p_1^2=2m(100k)$ $p_2^2=2m(400k)$ Fractional change in momentum $=\frac{p_2-p_1}{p_1}$ % change in momentum $=\frac{(p_2-p_1)}{p_1}\times 100$ $=\left(\frac{p_2}{p_1}-1\right)\times 100\left(\sqrt{\frac{4}{1}}-1\right)\times 100$ $=(2-1)\times 100$ $\therefore$ Increase in momentum = 100%