JEE MainPhysicsModerateWork Power & Energy

Initially spring is in natural length and system is at rest.
System is released when a particle of mass m collides vertically to
block B with velocity v and sticks to it. Find maximum extension in the
spring. (Pulley and string are ideal). (Given that
$m = 1kg,v = 10\sqrt{2}m/sec$ and spring constant $k = 10N/m$ )
( $g = 10m/s^{2})$

Show answer and solution

Just after collision velocity of (block B+
particle)

= \frac{v}{2}

At the time of maximum extension both A and B block will have same
velocity (say v')
Conserving angular momentum about centre of pulley of system of (A+B+
particle)

v^{'} = \frac{v}{4}

By energy conservation,

\frac{1}{2} \times 2m \times {(\frac{v}{2})}^{2} = 2 \times \frac{1}{2} \times 2m \times {(\frac{v}{4})}^{2} + \frac{1}{2} \times 10 \times A^{2} - 2mgA

On solving,
A=5m

Source: Grass-G / P-16 Physics paper 25 Nov First.docx

More JEE Main questions Chapter practice

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.

Scan a doubt View plans