Question

two identical blocks A and B each of mass 'm' resting on smooth floor are connected by light spring of natural length L and spring constant K , with the spring as it's natural length. a third identical block 'C' mass (m) moving with a speed V along the line joining A and B collides withA . the maximum compression in the SPRING is​

Answer 1

Answer:

v√m/k is the answer max compression

Answer 2

Answer:

$x=v\sqrt{\frac{m}{2k}}$

Explanation:

We are given that two identical blocks A and B.

Mass of each block=m

Block A and B are at rest.Therefore, their initial velocity is zero.

Length of spring=L

Spring constant=K

Mass of third block C=m

Speed of block C=v

We have to find the maximum compression in the spring.

According to momentum conservation  law

Momentum before collision=Momentum after collision

$mv=(m+m)v'$

$mv=2mv'$

$v'=\frac{v}{2}$

According to energy conservation law

$\frac{1}{2}mv^2=\frac{1}{2}(2m)v'^2+\frac{1}{2}kx^2$

$\frac{1}{2}mv^2=m(\frac{v}{2})^2+\frac{1}{2}kx^2$

$\frac{1}{2}mv^2=\frac{mv^2}{4}+\frac{1}{2}kx^2$

$\frac{1}{2}mv^2-\frac{1}{4}mv^2=\frac{1}{2}kx^2$

$\frac{1}{4}mv^2=\frac{1}{2}kx^2$

$x^2=\frac{1}{2}\frac{mv^2}{k}$

$x=v\sqrt{\frac{m}{2k}}$