First slide

Collisions

Question

Two identical blocks A and B, each of mass 'm' resting on smooth floor are connected by a light spring of natural length L and spring constant K, with the spring at its natural length. A third identical block 'C' (mass m) moving with a speed v along the line joining A and B collides with A. the maximum compression in the spring is

Moderate

A

$v \sqrt{\frac{m}{2 k}}$
B

$m \sqrt{\frac{v}{2 k}}$
C

$\sqrt{\frac{mv}{k}}$
D

$\frac{mv}{2 k}$

Solution

Initial momentum of the system (block C) = mv
After striking with A, the block C comes to rest and now both block A and B moves with velocity V, when compression in spring is maximum.
By the law of conservation of linear momentum
mv = (m + m) V $\Rightarrow V = \frac{v}{2}$
By the law of conservation of energy
K.E. of block C = K.E. of system + P.E. of system
$\frac{1}{2} {mv}^{2} = \frac{1}{2} (2 m) V^{2} + \frac{1}{2} {kx}^{2}$
$\Rightarrow \frac{1}{2} {mv}^{2} = \frac{1}{2} (2 m) {(\frac{v}{2})}^{2} + \frac{1}{2} {kx}^{2}$
$\Rightarrow {kx}^{2} = \frac{1}{2} {mv}^{2}$
$\Rightarrow x = v \sqrt{\frac{m}{2 k}}$

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