Question

a block of mass 2 kg is moving on a frictionless horizontal surfaces with velocity of 1 m/s towards another block of equal mass kept at rest .the spring constant of the spring fixed at one end us 100 n/m . find the maximum cimpression in the spring.

Answer 1

Mass of each block MA and MB = 2kg.
Initial velocity of the 1st block, (V) = 1m/s
VA = 1 m/s, VB = 0m/s
Spring constant of the spring = 100 N/m.
The block A strikes the spring with a velocity 1m/s/
After the collision, it’s velocity decreases continuously and at a instant the whole system (Block A + the compound spring + Block B) move together with a common velocity.
Let that velocity be V.

Using conservation of energy,

(1/2) MAVA 2 + (1/2)MBVB 2 = (1/2)MAv2 + (1/2)MBv2 + (1/2)kx2

(1/2) × 2(1)2 + 0 = (1/2) × 2× v2 + (1/2) × 2 × v2 + (1/2) x2 × 100

(Where x = max. compression of spring)

⇒1 = 2v2 + 50x2 …(1)

As there is no external force in the horizontal direction, the momentum should be conserved.

Hope this helps u Please make me as a brainliest