Squeezing A Spring

Classical Mechanics Level 2

The block of mass m m moving on a frictionless horizontal surface collides with the spring of spring constant k k , compresses it by length L , L, and then rebounds. What is the maximum momentum of the block after it loses contact with the spring?

L m k L \sqrt{mk} m L 2 \frac{mL}2 m L 2 \frac{m}{L^2} m L k \sqrt{m L k}

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Syed Amaan Haider by Syed Amaan Haider , 23, India

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2 solutions

Andrew Ellinor
Feb 11, 2016

As the block moves across the surface with velocity v , v, it has kinetic energy of 1 2 m v 2 . \frac{1}{2}mv^2. A spring's potential energy is 1 2 k x 2 , \frac{1}{2}kx^2, where x x is its displacement from equilibrium. So, when the spring is compressed by the block, it reaches a potential energy of 1 2 k L 2 . \frac{1}{2}kL^2.

By conservation of energy , 1 2 m v 2 = 1 2 k L 2 ⟶ v = k L 2 m \frac{1}{2}mv^2 = \frac{1}{2}kL^2 \longrightarrow v = \sqrt{\frac{kL^2}{m}}

Therefore, the momentum of the given object is

m v = m k L 2 m = L m k . mv = m \sqrt{\frac{kL^2}{m}} = \boxed{L\sqrt{mk}}.

9 Helpful 1 Interesting 1 Brilliant 0 Confused

nice job bro

Haroon Marwat - 5 years, 4 months ago

I made it based only on dimensional analysis considerations...

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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