MCQ
If momentum is increased by $20 \%$, then K.E. increases by
- ✓$44 \%$
- B$55 \%$
- C$66 \%$
- D$77 \%$
✓
Answer
Correct option: A.
$44 \%$
(a)$E=\frac{P^2}{2 m}$ . If $m$ is constant then
$E\propto P^2 \Rightarrow \frac{E_2}{E_1}=\left(\frac{P_2}{P_1}\right)^2=\left(\frac{1.2P}{P}\right)^2=1.44 $
$\Rightarrow E_2=1.44 E_1=E_1+0.44 E_1 $
$E_2=E_1+44 \% \text { of }E_1$
i.e.the kinetic energy will increase by $44 \%$
$E\propto P^2 \Rightarrow \frac{E_2}{E_1}=\left(\frac{P_2}{P_1}\right)^2=\left(\frac{1.2P}{P}\right)^2=1.44 $
$\Rightarrow E_2=1.44 E_1=E_1+0.44 E_1 $
$E_2=E_1+44 \% \text { of }E_1$
i.e.the kinetic energy will increase by $44 \%$
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