# Centre of mass problems

1. Dec 8, 2013

### MathewsMD

21. A 2.0-kg block is attached to one end of a spring with a spring constant of 100N/m and a 4.0-kg block is attached to the other end. The blocks are placed on a horizontal frictionless surface and set into motion. At one instant the 2.0-kg block is observed to be traveling to the right with a speed of 0.50 m/s and the 4.0-kg block is observed to be traveling to the left with a speed of 0.30m/s. Since the only forces on the blocks are the force of gravity, the normal force of the surface, and the force of the spring, we conclude that:
A. the spring is compressed at the time of the observation
B. the spring is not compressed at the time of observation
C. the motion was started with the masses at rest
D. the motion was started with at least one of masses moving
E. the motion was started by compressing the spring

22. A 2.0-kg mass is attached to one end of a spring with a spring constant of 100 N/m and a 4.0-kg mass is attached to the other end. The masses are placed on a horizontal frictionless surface and the spring is compressed 10 cm. The spring is then released with the masses at rest and the masses oscillate. When the spring has its equilibrium length for the first time the 2.0-kg mass has a speed of 0.36 m/s. The mechanical energy that has been lost to the instant is:
A. zero
B. 0.31 J
C. 0.61 J
D. 0.81 J
E. 1.2J

Hi, I'm just having trouble understanding these two questions.

For 21, why cannot the motion have started by compressing the spring? I believe I understand why there must have been at least one mass moving, but I don't know why it cannot have began by an initial compression?

For 22, ΔM = Mf - Mi
Based on linear momentum, if pi = 0, then pf = 0; (2 kg)(0.36 m/s)/(4 kg) = v2 = 0.18 m/s
Mf = Kf since U = 0 J b/c there is no potential energy in the equilibrium position.
Mf = 1/2 [ (2kg)(0.36 m/s)2 + (4kg)(0.18 m/s)2 ] = 0.19 J

Now, the initial mechanical energy, Mi, equals just the spring potential energy, 1/2kx2 and that is 5 J

So, ΔM = Mf - Mi = 0.19 J - 5 J = -4.81 J and that is the amount of mechanical energy lost.

Please correct me where i've gone wrong and if you can, offer an explanation so I can follow along. :)

Ans 21: D <-- Highlight to see answer more clearly
Ans 22: B <-- Highlight to see answer more clearly

2. Dec 8, 2013

### MathewsMD

Also:
40. Cart A, with a mass of 0.20kg, travels on a horizontal air track at 3.0m/s and hits cart B, which has a mass of 0.40 kg and is initially traveling away from A at 2.0 m/s. After the collision the center of mass of the two cart system has a speed of:
A. zero
B. 0.33 m/s
C. 2.3 m/s
D. 2.5 m/s
E. 5.0 m/s
ans: B

These answers seem really simple yet I keep getting different answers. I solved for the centre of mass speed by doing: vcom = (mAvA + mBvB)/mtotal and vB is 2.0 m/s relative to A, so it is 5 m/s. Well, I did this and found 4.33 m/s...it's not even an option. I can't seem to get the answer of B.

3. Dec 8, 2013

### MathewsMD

Sorry for the questions. I have all the answers but am having trouble finding the solutions:

69. A 75-kg man is riding in a 30-kg cart at 2.0m/s. He jumps o in such a way as to land on the ground with no horizontal velocity. The resulting change in speed of the cart is:

A. zero
B. 2.0 m/s
C. 3.0 m/s
D. 5.0 m/s
E. 7.0 m/s

Ans: D

I assumed it was a simple linear momentum question and I did, [(m + M)vi]/m = vf since m is the mass of the cart while M is the man. Also, the questions says the man has no horizontal velocity, and this means momentum in the x-direction is 0 for the man, thus, the formula I just stated should work...the answer should be 7.0 m/s according to my attempt but that is once again sadly incorrect.