Question

If kinetic energy of a body is increased by $300$ % then percentage change in momentum will be

• A. 1
• B. 1.5
• C. 2.65
• D. 0.732

Answer 1

Answer: (A)

1

Answer 2

$P _{1} =\sqrt{2 mE _{1}} ; P _{2}=\sqrt{2 mE _{2}}$
$=\sqrt{2 m \left( E _{1}+\frac{300}{100} E _{1}\right)}=\sqrt{2 m \left(4 E _{1}\right)}=2 P _{1}$
$\%$ change $=\frac{P_{2}-P_{1}}{P_{1}} \times 100=\frac{2 P_{2}-P_{1}}{P_{1}} \times 100$
$=100 \%$