Centre Of Mass Ques 66
- In a collinear collision, a particle with an initial speed $v _0$ strikes a stationary particle of the same mass. If the final total kinetic energy is $50 %$ greater than the original kinetic energy, the magnitude of the relative velocity between the two particles after collision, is
(2018 Main)
(a) $\frac{v _0}{\sqrt{2}}$
(b) $\frac{v _0}{4}$
(c) $\sqrt{2} v _0$
(d) $\frac{v _0}{2}$
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Answer:
Correct Answer: 66.(c)
Solution:
- From conservation of linear momentum,
$$ \begin{aligned} & m v _0+0=m v _1+m v _2 \\ & v _0=v _1+v _2 \\ & \text { Further, } \quad K _f=\frac{3}{2} K \text {, } \\ & \therefore \quad \frac{3}{2} \frac{1}{2} m v _0^{2}=\frac{1}{2} m v _1^{2}+\frac{1}{2} m v _2^{2} \\ & \bigcirc _m \bigcirc _m^{v _0} \overbrace{m}^{\text {Rest }} \bigcirc _m^{v _1} \\ & \text { Before collision After collision } \\ & \frac{3}{2} v _0^{2}=v _1^{2}+v _2^{2} \end{aligned} $$
Solving Eqs. (i) and (ii),
$$ \begin{gathered} v _1=\frac{v _0}{2}(1+\sqrt{2}) \\ v _2=\frac{v _0}{2}(1-\sqrt{2}) \\ v _{\text {rel }}=v _1-v _2 \\ \frac{v _0}{2}[1+\sqrt{2}-1+\sqrt{2}]=\frac{v _0}{2} \times 2 \sqrt{2}=\sqrt{2} v _0 \end{gathered} $$
BETA
