Question

Blocks A (mass 4.50 kg) and B (mass 15.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 1.00 m/s.The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line. Let +x be thedirection of the initial motion of A

Find the maximum energy stored in the spring bumpers and the velocity of each block at that time.
Find the maximum energy.

Find the velocity of A &B

Find the velocity of each block after they have moved apart.
Find the velocity of A &B

There are problems like this but i don't understand how and why they set it up in that way.

Answer 1

the conservation of momentum ,
m1u1+0 =(m1+m2)V
V = 4.5/19.5 = 0.23 m/s
the velocity of A&B is 0.23 m/s.
the maximum energy store is equals to the change in K.E of the system.
ΔK.E =1/2*4.5*1.0- 1/2*19.5*(0.23)^2
= 2.25-0.519
= 1.730 J
the maximum energy stored spring bumpers is 1.730 J.
the maximum energy = 1.730 J
According to conservation of momentum and elastic collision,
m1u1+0 = m1v1 + m2v2 ---------------- (1)
u1 = v2-v1 ------------------ (2)
From equation (1)&(2)
m1u1 = m1(v2-u1) + m2v2
2m1u1 = v2(m1+m2)
2*4.50*1.0 = v2(19.5)
v2 = 0.461 m/s
and
v1 = -0.538 m/s