Question

If momentum decreases by 20%, KE will decrease by

• A. 40%
• B. 18%
• C. 36%
• D. 8%

Answer 1

Answer: (C)

36%

Let initial momentum of the body be $p$ and its mass be $m$.

Thus initial kinetic energy $K.E_i=\frac{p^2}{2m}$

Now the momentum decreases by 20%, thus final momentum $p_f=0.8p$

Final kinetic energy $K.E_f=\frac{(p_f{)}^2}{2m}$

$\therefore$ $K.E_f=\frac{(0.8p{)}^2}{2m}=0.64K.E_i$

Percentage change in kinetic energy $\frac{\Delta K.E}{K.E_i}\times 100=\frac{K.E_i-0.64K.E_i}{K.E_i}\times 100$

$⟹$ $\frac{\Delta K.E}{K.E_i}\times 100=0.36\times 100=36$ %

Thus kinetic energy decreases by $36$ %.