Question 236: Centre of Mass and Rotational Motion

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

Accepted Answer

Elastic energy stored in the spring = ${1 \over 2}k{L^2}$
And kinetic energy of the block = ${1 \over 2}M{v^2}$
$	herefore$ ${1 \over 2}M{v^2} = {1 \over 2}k{L^2}$
$ \Rightarrow v = \sqrt {{k \over M}} .L$
$	herefore$ Momentum $ = M 	imes v = M 	imes \sqrt {{k \over M}} .L = \sqrt {kM} .L$

Answer

${{k{L^2}} \over {2M}}$

Answer

$\sqrt {Mk} \,\,L$

Answer

${{M{L^2}} \over k}$

Answer

Zero

Centre of Mass and Rotational Motion

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

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