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Perfectly Elastic Collision

Two blocks A ...

Question

Two blocks A and B of masses $m_{1}$ and $m_{2}$ respectively are connected with each other by a spring of force constant k as shown in figure. Block are pulled away from each other by $x_{0}$ and then released. when spring is in its natural length and blocks are moving towards each other, another block of mass m moving with velocity $v_{0}$ (towards right) collides with A and gets stuck to it. Neglect friction Let $v_{1}$ and $v_{2}$ be the velocities of the blocks A and B respectively just before collision. Let $v_{c m}$ be velocity of center of mass of the system; after collision.
Column -I Column- II
A) Initial kinetic
energy of system
p) $(m_{1} + m_{2} + m) v_{c m}$
B) Initial momentum
of system
q) $\frac{1}{2} (m + m_{1} + m_{2}) v_{c m}^{2}$
C) Final momentum
of system
r) $\frac{1}{2} m v_{0}^{2} + \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$
D) Final kinetic energy
of system
s) None of these

A

A-r

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B

B-p

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C

C-p

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D

D-q

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Solution

The correct options are
A A-r
B B-p
C C-p
D D-q
According to given Condiation:
Initial kinetic energy of the system is equal to the sum of individual kinetic energies= $\frac{1}{2} m v_{0}^{2} + \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$

Since no external force acts on the system in the horizontal direction, the momentum of the system in the direction is conserved.
Hence initial momentum=final momentum= $(m + m_{1} + m_{2}) v_{c m}$
All the particles finally attain an equivalent center of mass velocity after a collision and hence the final kinetic energy of the system can be expressed as

$\frac{1}{2} (m + m_{1} + m_{2}) v_{c m}^{2}$

All option are correct

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