Question

Two blocks of masses 5 kg and 2 kg are connected by a spring of negligible mass placed on a frictionless horizontal surface an impulse provided a velocity of 7 metre per second of the heavier block in the direction of light of the velocity of the centre of masses

Answer 1

M1 = 5 kg M2= 2 kg

So, impulse is provided to hevier mass i.e; 5 kg block

M1V1 + M2V2 / M1+ M2

5 x 7 + 2x 0 / 7

35/7 = 5 m/s

Answer 2

Here's the answer

Use the basic formula m1v1+m2v2/m1+m2

Then impulse is only given to the heavier block i.e of 5 kg

And lighter block will have 0 impulse.

=5×7+2×0/5+2== 5m/s