First slide

The steps in collision

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 elastically , 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

$\sqrt{\frac{mv}{2 k}}$

Solution

After striking with A, the block C comes to rest and block A moves with velocity v. When compression in spring is
maximum, both A and B will be moving with common velocity v.
From law of conservation of linear momentum,
$mv = (m + m) V \Rightarrow V = \frac{v}{2}$ …(i)
From law of conservation of energy,
KE of block C = KE of system + PE of system
$\frac{1}{2} {mv}^{2} = \frac{1}{2} (2 m) v^{2} + \frac{1}{2} {kx}^{2}$
$\Rightarrow \frac{1}{2} {mv}^{2} = \frac{1}{2} (2 m) {(\frac{v}{2})}^{2} + \frac{1}{2} {kx}^{2}$ [from Eq. (i)]
$\Rightarrow {kx}^{2} = \frac{1}{2} {mv}^{2} \Rightarrow x = v \sqrt{\frac{m}{2 k}}$

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