Question

319. Two identical blocks A and B, each of mass m
resting on smooth floor are connected by a light
spring of natural length Land spring constant K,
with the spring in its natural length. A third
identical block of mass m, moving with a speed
v along the line joining A and B collides with A.
The maximum compression in the spring is
(a) v
V
(b) m
2 K
2K
Vek
mv
(c)
(d)
2 K
Mironto​

Answer 1

Explanation:

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Answer 2

Answer:

At the instant when the third block collides with one of the block attached to the spring, since the spring is in natural length which means it is exerting no force, also there's no friction present.

So, we can conserve linear momentum of the system of third block and one of the block attached to the spring.

There arise two cases: 1) Elastic collision

2) Inelastic collision.

I'll do elastic one for you .

1) Conserving linear momentum,

mv = mu

implies, u = v

Now,

Conserving energy of the spring- two block system.

1/2mu^2 + 1/2m(0)^2 = 1/2kx^2

which gives,

x (max) = u√(m/k) = v√(m/k)

If , it had been Inelastic collision, then just conservation of momentum equation would be changed.

That is,

mv = mu+mu

or,

u = v/2

Clearly, more spring will be compressed in case of elastic collision.

Hope this helps you !