Centre Of Mass Ques 38
- Two particles of masses $m _1$ and $m _2$ in projectile motion have velocities $\mathbf{v} _1$ and $\mathbf{v} _2$ respectively at time $t=0$. They collide at time $t _0$. Their velocities become $\mathbf{v} _1^{\prime}$ and $\mathbf{v} _2^{\prime}$ at time $2 t _0$ while still moving in air. The value of
$\left|\left(m _1 \mathbf{v} _1^{\prime}+m _2 \mathbf{v} _2^{\prime}\right)-\left(m _1 \mathbf{v} _1+m _2 \mathbf{v} _2\right)\right|$ is
(2001, S)
(a) zero
(b) $\left(m _1+m _2\right) g t _0$
(c) $2\left(m _1+m _2\right) g t _0$
(d) $\frac{1}{2}\left(m _1+m _2\right) g t _0$
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Answer:
Correct Answer: 38.(c)
Solution:
Formula:
Motion of Centre of Mass and Conservation of Momentum:
- $\left|\left(m _1 \mathbf{v} _1^{\prime}+m _2 \mathbf{v} _2^{\prime}\right)-\left(m _1 \mathbf{v} _1+m _2 \mathbf{v} _2\right)\right|$
$=\mid$ change in momentum of the two particles $\mid$
$=\mid$ External force on the system $\mid \times$ time interval
$$ =\left(m _1+m _2\right) g\left(2 t _0\right)=2\left(m _1+m _2\right) g t _0 $$
BETA
