Question

A block of mass 4m moves on a flat horizontal surface at speed v, and strikes another block of mass m, that is attached via a horizontal spring to another block of mass 2m. The two masses are stationary.  The spring constant is k. All collisions are completely inelastic and there is no friction anywhere.
What is the velocity of the center of mass when the spring is ¼ the way to max compression.

Answer 1

See the situations happened :

Given : spring constant = k.

assuming Maximum compression = A

x = A/4 = 1/4 of maximum compression.

When 4m collides with m,

applying conservation of linear momentum :

4m x v = 5m x v' [since, collision is completely inelastic].

=> v' = 4v/5.

When compression is maximum, the velocities of the bodies will be same,

applying conservation of linear momentum :

5m x v' = 7m x v''

=> v'' = 5v'/7 = 4v/7.

Also, here we can apply conservation of energy :

------- (i)

When compression (x) is 1/4 of maximum, the velocity of 5m is v1 and velocity of 2m is v2.

applying conservation of momentum :

(5m x v' )= (5m x v1) + (2m x v2)

=> 4v = 5v1 + 2v2 ----- (ii)

applying conservation of energy :

[puting the value from equa. (i)]

-------(iii)

On solving equa. (ii) and (iii) ,

we get the acceptable value of v1 and v2 during compression:

v1 = 0.79 m/s.

v2 = 0.02 m/s.

Hence, vel of center of mass

Also, we have another set of value of v1 and v2 during relaxation from maximum compression and reached x = A/4,

v1 = 0.35 m/s

v2 = 1.12 m/s.

Hence , vel of center of mass

The ans remain same because, there is no external force acting.

Alternative Method for the problem :

As there is no external force acting at any point on the system of three bodies,

the velocity of center of mass will remain same always at all positions.

Hence, vel of center of mass before collision