Question

the block of mass M moving on the frictionless horizontal surface college with spring constant compressors tips by length the maximum momentum of the block after collision is

Answer 1

$\frac{1}{2} \times k {l}^{2} = \frac{ {p}^{2} }{2m} \\ hence \\ p = \sqrt{mk} \times l$

Answer 2

Answer:

Maximum momentum is$P=x\sqrt{mk}$

Explanation:

Since we know that the energy of the spring is given by

$E=\dfrac{kx^{2} }{2}$.......................(1)

We also know the momentum of the body,It is the total quantity of motion performed by the body.

$P=mv$......................................(2)

The kinetic energy of the block is

$K.E.=\dfrac{mv^{2}}{2}$...............................(3)

From (2) and (3) we get,

$K.E=\dfrac{P^{2} }{2m}$..........................(4)

Since equation (1) and (4) represents the energy,so we are equating both the equation

$\dfrac{kx^{2}}{2}=\dfrac{P^{2} }{2m}$

So,maximum momentum is$P=x\sqrt{mk}$