# Linear momentum- moving blocks and attatched spring

1. Mar 23, 2009

### rmunoz

1. The problem statement, all variables and given/known data
In the figure below, block 1 (mass 1.6 kg) is moving rightward at 12 m/s and block 2 (mass 4.8 kg) is moving rightward at 2.8 m/s. The surface is frictionless, and a spring with spring constant of 1100 N/m is fixed to block 2. When the blocks collide, the compression of the spring is maximum at the instant the blocks have the same velocity. Find the maximum compression.
http://www.webassign.net/halliday8e/pc/halliday8019c09/halliday8019c09-fig-0065.htm [Broken]

2. Relevant equations
m1(i)v1(i) + m2(i)v2(i) = m1(f)v1(f) + m2(f)v2(f)
p=mv
KE=1/2mv$$^{2}$$
Work-spring=1/2kx$$^{2}$$

3. The attempt at a solution
Momentum before collision:
P1+P2= Psys
m1v1 + m2v2 = Psys
1.6kg*12m/s +4.8kg*2.8m/s= 19.2 kg*m/s + 13.44 kg*m/s = 32.64 kg*m/s

work done by spring on object 1 = change in object 1's kinetic energy ?

Momentum after collision of object 1 will be negative, and momentum of object 2 will have increased in the positive direction.

Questions:
1) how do i relate the compression of the spring to the linear momentum of the system?
2) how do i tell weather or not linear momentum is conserved (in general, not just in the scope of this problem)
3) how can i relate the change in kinetic energy due to the spring to the change in the momentum of object 1 and 2? (or is that at all how i am supposed to approach this problem)

Last edited by a moderator: May 4, 2017
2. Mar 23, 2009

### lanedance

Hi rumonz

can't see the picture but i assume block 1 impacts the spring on block 2?

1) energy of a whole system is always conserved (by a whole system I mean everything that is interacting) - look at the energy in your system
2) linear momentum of a whole system is always conserved
3) You're on the right track... Kinetic energy and momentum are both functions of velocity. And the question asks you the compression of the spring when both masess are moving at the same velocity. Can you solve for this velocity?

3. Mar 24, 2009

### rmunoz

So i solved for the velocity by the following:
Psys=32.64kg*ms
V1=V2

m1v1(comp) + m2v2(comp) = 32.64kg*m/s
v1comp(m1 + m2)= 32.64kg*m/s
v1comp=[32.64kg*m/s]/[6.4kg]
v1comp= 5.1 m/s

Now im stuck again. I understand that $$\Delta$$KE= work and I think that has some relation to the change in kinetic energy of mass 1 by the spring system aka 1/2Kx^2. I want to say that the kinetic energy of mass 1 can be described by adding -1/2Kx^2 (which should be a negative value since its pushing away from object one) to the initial kinetic energy of mass 1. This is what i did to solve for the compression:

1/2m1vo^2 - 1/2kx^2 = KE m1 at fullcompression= Pcompm1(vcomp)(1/2)
yielding:

x^2 (compression distance) = [(m1vo^2)-Pcomp_m1*(vcomp)]/(K)

x= $$\sqrt{[230.4kg*m^2/s^2 - 41.616kg* m^2/s^2]/1100}$$
x= .4143 m

What did i do wrong? I suspect I missed something quite large, because the numbers are really not coming out too friendly.

help?

4. Mar 24, 2009

### lanedance

I'm not too sure what you did in the 2nd part

the total energy of the system will be conserved at all time in this case, that amounts to:

$$E = KE + PE = (\frac{1}{2} m_1 v_1^2+ \frac{1}{2} m_2 v_2^2) + \frac{1}{2}k x^2$$