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Question

(d)

In collision, both kinetic energy and linear momentum are conserved.

So, we have

$\frac{1}{2} m u^2=\frac{1}{2} m v_1^2+\frac{1}{2}(2 m)

Vedclass · Accepted Answer

$\frac{h}{9}$

Vedclass · Answer

(d)

In collision, both kinetic energy and linear momentum are conserved.

So, we have

$\frac{1}{2} m u^2=\frac{1}{2} m v_1^2+\frac{1}{2}(2 m) v_2^2$

and $m u =2 m v_2-m v_1$

$\Rightarrow u^2 =v_1^2+2 v_2^2 \ldots (i)$

and $ u=2 v_2-v_1 \quad \ldots (ii)$

Substituting for $v_2$ from Eq. $(ii)$ in Eq. $(i)$,

we get

$u^2=v_1^2+2\left(\frac{u+v_1}{2}\right)^2$

$\Rightarrow2 u^2=2 v_1^2+u^2+v_1^2+2 u v_1$

$\Rightarrow u^2-v_1^2=2 v_1\left(u+v_1\right)$

$\Rightarrow u-v_1 =2 v_1$

$\Rightarrow v_1=\frac{u}{3}$

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