A block of mass 2.0 kg is moving on a frictionless horizontal surface with a velocity of 1.0 m/s figure towards another block of equal mass kept at rest. The spring constant of the spring fixed at one end is 100 N/m. Find the maximum compression of the spring.

Explanation

Mass of each block `M_B=2kg`
Initial velocity of the 1st block
`v=1m/sec`
`V_a=1m/sec`
`V_b=0m/sec`
Spring constantof the spring =100N/m

The block A strikes the spring with a velocity 1m/sec.
After teh colision, its velocilty deceases continuously adn at an instant the whole system(Block A+the compound spring +Block B) moves together wilth a common velocity.
Let teh velocity be V.
Using conservation of energy,.
`(1/2)M_AV_A^2+(1/2)M_bV_B^2`
`=(1/2)M_AV^2+(1/2)M_BV^2+(1/2)kx^2`
`(1/2)xx2(1)^2+0`
`=(1/2)+(1/2)xv^2+(1/2)x^2xx100`
`(when x=maximum compresion of spring)
`rarr `1-2v^2=50x^2`..........i
`As there is no exterN/Al force in the horizontal direction, the momentum should be conserved.
`rarr M_AV_A+M_BV_B=(M_A+M_B)V`
`rarr 2xx1=4xxV`
`rarr V=(1/2)m/sec` .....ii
putting in equationn i
`1=2xx(1/4)+50x^2`
`rarr (1/4)=50x^2`
`rarr (1/4)=50x^2`
`rarr x^2=1/100m^2`
`rarr x=(1/10)m`
`=0.1m=10cm`