JEE Main 2005

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:

Block-spring collision on frictionless surface

![Block-spring collision on frictionless surface](https://www.edvaya.com/jee/physics/q10-block-spring-frictionless-2005.png) A). $\dfrac{kL^2}{2M}$ B). $\sqrt{Mk}\,L$ C). $\dfrac{ML^2}{k}$ D). Zero

Edvaya · Accepted Answer

The correct answer is $\sqrt{Mk}\,L$. The block has maximum momentum at the moment just before hitting the spring (before any compression), as the spring force will subsequently decelerate it.
At maximum compression $L$, all kinetic energy converts to spring potential energy:
$\dfrac{1}{2}Mv_0^2 = \dfrac{1}{2}kL^2 \Rightarrow Mv_0^2 = kL^2$
Maximum momentum $p = Mv_0$:
$p^2 = M^2v_0^2 = M \cdot kL^2$
$p = L\sqrt{Mk}$

Edvaya Target

Edvaya Aspire

Take a Free Mock Test

Target Subjects

Aspire Subjects

Subject Mastery

Physics > Mechanics > Work Energy And Power

Keep Practicing

Current Question

Community Stats

Track Your Progress

Edvaya Target

We Value Your Privacy