A block of mass 2.0 kg is moving on a frictionless horizontal surface with a velocity of 1.0 m/s figure towards abnother block of equal mass kept at rest. The spring constant of the spring fixed at one end is 100 The spring constant of the spring fixed at one end is 10 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`