Question

When the momentum of a body increases by $100$% its kinetic energy increases by:

• A. $400$%
• B. $100$%
• C. $none$
• D. $300$%

Answer 1

Answer: (D)

$300$%

Let the initial momentum is $p$ and initial kinetic energy is K.

Thus, $K=\frac{p^2}{2m}$

When momentum increases by 100 % , the momentum becomes $p^{\prime }=p+(p\times 100/100)=p+p=2p$

Thus, kinetic energy $K^{\prime }=\frac{p^{\prime 2}}{2m}=\frac{(2p{)}^2}{2m}=4K$

or $K^{\prime }/K=4$

Hence, the % increase in kinetic energy $=\frac{K^{\prime }-K}{K}\times 100=(K^{\prime }/K-1)\times 100=(4-1)100=300$ %