Question

Two bodies $P$ and $Q$ having masses in the ratio $1:4$ and having K.E in the ratio of $4:1$ then the ratio of linear momentum of $P$ and $Q$ is

• A. $1:4$
• B. $1:2$
• C. $1:16$
• D. $1:1$

Answer 1

The correct option is D $1:1$

Given$:$

$\frac{m_A}{m_B}=\frac{1}{4}$

$\frac{\frac{1}{2}\times m_A\times {\left(v_A\right)}^2}{\frac{1}{2}\times m_B\times {\left(v_B\right)}^2}=\frac{4}{1}$

$\Rightarrow \left(\frac{m_A}{m_B}\right){\left(\frac{v_A}{v_B}\right)}^2=\frac{4}{1}$

$\Rightarrow \frac{v_A}{v_B}=\sqrt{\frac{16}{1}}=4:1$

So$,$ $\frac{m_A{v}_A}{m_B{v}_B}=\left(\frac{m_A}{m_B}\right)\left(\frac{v_A}{v_B}\right)=\frac{1}{4}\times \frac{4}{1}=1:1$

Hence,

option $\left(D\right)$ is correct answer.