First slide

Conservation of mechanical energy

Question

The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant K and compresses it by length L. The maximum momentum of the block after collision is

Moderate

A

Zero
B

$\frac{{ML}^{2}}{K}$
C

$\sqrt{MK} L$
D

$\frac{{KL}^{2}}{2 M}$

Solution

When block of mass M collides with the spring its kinetic energy gets converted into elastic potential energy of the spring.
From the law of conservation of energy
$\frac{1}{2} {Mv}^{2} = \frac{1}{2} {KL}^{2} ∴ v = \sqrt{\frac{K}{M}} L$
Where v is the velocity of block by which it collides with spring. So, its maximum momentum
$P = Mv = M \sqrt{\frac{K}{M}} L = \sqrt{MK} L$
After collision the block will rebound with same linear momentum.

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