A block of mass m1 = 2 kg slides along a frictionless table with a velocity of +10 m/s.?

Question

Directly in front of it, and moving with a velocity of +3 m/s, is a block of mass m2 = 9 kg. A massless spring with spring constant R = 1120 N/m is attached to the second block as in Figure 8-59.

a) Before m1 runs into the spring, what is the velocity of the center of mass of the system?

(b) After the collision, the spring is compressed by a maximum amount x. What is the value of x?

(c) The blocks will eventually separate again. What is the final velocity of each block measured in the reference frame of the table?

Answer 1

a) 2(10) + 9(3) = 11Vcm

Vcm = 4.2 m/s

b) When the spring is compressed at its maximum amount, the velocity of the two blocks will be the same as each other. It will also be the same as the answer to a. The kinetic energy that is lost in the collision must be stored in the spring, so...

1/2(2)10^2 + 1/2(9) 3^2 = 1/2(1120)x^2 + 1/2(11)(4.2)^2
KE1Before + KE2Before = PEspring + KEafter

Solve this for x, and you'll have the maximum compression of the spring during the collision.

c) I think they want us to assume the collision is perfectly elastic, in which case, you use momentum conservation

2(10) + 9(3) = 2(V1A) + 9(V2A)

AND the elastic collision relative velocity equation:

V2B - V1B = V1A - V2A
3-10 = V1A - V2A

Now you have two equations and two unknowns.