Question

If $K.E.$ body is increased by $100$. Then change ${}^{\prime }{P}^{\prime }$:

• A. $50\%$
• B. $41\%$
• C. $10\%$
• D. $20\%$

Answer 1

Answer: (B)

$41\%$

We have $KE=\frac{1}{2}{MV}^2⟶\left(1\right)$

& linear momentum $P=MV⟶\left(2\right)$

Equation $\left(1\right)$ & $\left(2\right)$ $\Rightarrow KE=\frac{P^2}{2M}$

Given $KE=2K$

Let new momentum be $P^{\ast }$

$\therefore$ $\frac{K^{\ast }}{K}={\left(\frac{P^{\ast }}{P}\right)}^2$

$\frac{2K}{K}={\left(\frac{P^{\ast }}{P}\right)}^2\Rightarrow {2P}^2=P^{{\ast }^2}$

or $P^{\ast }=\sqrt{2}P$

The momentum increases by $\sqrt{2}=1.41$ times

$=\left(\frac{1.41-1}{1}\right)\times 100=41$%