# Springs w/ Collisions

1. Nov 16, 2007

### bphysics

1. The problem statement, all variables and given/known data

A block of mass M is resting on a horizontal, frictionless table and is attached as shown above to a relaxed spring of spring constant k. A second block of mass 2 M and initial speed v0 collides with and sticks to the first block.

Develop expressions for the following quantities in terms of M, K, and v0.

a) v, the speed of the blocks after impact

b) x, the max distance the spring is compressed

c) T, the period of the subsequent simple harmonic motion

2. Relevant equations

- M1V1 + M2V2 = (M1 + M2)(Vf)

- (1/2)(k)(x^2)

- F = -kx

- (1/2)(m)(v^2)

3. The attempt at a solution

a)

M1V1 + M2V2 = (M1 + M2)(Vf)

(2M)(V0) + (M)(0) = (2M + M)(x)

(2M)(V0) = (3M)(Vf)

Vf = (2/3)(V0)

b)

x = max distance spring is compressed

spring constant = k

PE of spring = (1/2)(k)(x^2)

Max distance = PE at MAX, KE = 0

Logic process: We know that v = Vf is solved above, set PE = KE, since complete transfer occurs

(1/2)(m)(v^2) = (1/2)(k)(x^2)

(m)((2/3)(V0))^2 = (k)(x^2)

sqrt(((m)((2/3)(V0))^2) / k) = x

c)

No clue, unknown how to solve this (help???)

2. Nov 16, 2007

### Shooting Star

T = 2*pi*sqrt(m/k) for SHM.

3. Nov 16, 2007

### bphysics

I'm sorry -- is the rest of my work correct for parts A and B?

I "assumed" knowng myself that I would be wrong up there somewhere -- I guess not this time :)

4. Nov 16, 2007

### Shooting Star

The rest of your solution is quite correct. Good work.

5. Feb 5, 2008

### Swedishfish

Can I assume that since there is nothing acting to dampen the SHM that the X found in part b.) for max compression is also the amplitude?

6. Feb 6, 2008

### Shooting Star

I noticed an error here today. The 2nd eqn should be (3m)((2/3)(V0))^2 = (k)(x^2). The x will change.

No. There is no mention of damping in the problem, and we'll assume that the motion is undamped, and max compression is also the amplitude.