Question 235: 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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