# Homework Help: Qns On Momentum

1. Jun 12, 2006

### Delzac

here is how it goes,

A block of mass $$m_1 = 2.0kg$$ slides along a frictionless table with a speed of 10m/s. Directly in front of it, and moving in the same direction, is a block of mass $$m_2 = 5.0kg$$ moving at 3.0m/s. A massless spring with a spring constant k=1120N/m is attached to the backside of $$m_2$$. when the block collide, what is the maximum compression of the spring? Assume that the spring does not bend and always obeys Hooke's law.

Ans: 0.25 m

Let K= spring constant
e= extension

Here's how i tried to do,

i calculate that relative speed of $$m_1$$ to $$m_2$$ = 7m/s

then using this value calculate the KE and equate it to E= $$\frac{1}{2} k e^2$$.

but the ans is wrong.

Then,

i used the conservation of inelastic collision formula,

$$m_1 u_1 + m_2 u_2 = m_(1+2) V$$

used the velocity V and calculate the KE and Equate is into E= $$\frac{1}{2} k e^2$$ but didn't work.

how should i solve this prob?

Last edited: Jun 12, 2006
2. Jun 12, 2006

### DaMastaofFisix

The first step in solving this problem is that your relative appreoach is the right framework to be working in. Remember that because the spring is in contact or rather a part of the problem, conservation of both momentum and energy is applicable. The catch is knowing when each law applies, for instance where momentum is conserved and how energy is lost.

3. Jun 12, 2006

### DaMastaofFisix

And ah yes, be consistent with your algebraic setup and labeling of variables.

4. Jun 12, 2006

### Delzac

is the collision elastic or inelastic, since the spring is going to be compress

5. Jun 12, 2006

### Delzac

say i take the collision to be elastic :

$$v_2f = ( \frac{2m_1}{m_1 + m_2}) v_1i$$ given $$v_2i$$ is zero.

Final Velocity of $$m_2$$ = 4

using

kinetic energy = elastic potential energy

$$\frac{1}{2} m v^2$$ = $$\frac{1}{2} k e^2$$
$$\frac{1}{2} (5) (4)^2$$ = $$\frac{1}{2} (1120) e^2$$

the extension i obtain is 0.267m, which is wrong.

what did i do incorretly?

6. Jun 13, 2006

### arunbg

Although, I haven't gone through your complete solution, I can outline an approach.
1)Conservation of energy can be used.
2)Conservation of momentum for the complete system holds.
3)Once the masses collide, both move with a common velocity .
This velocity can be obtained through conservation of momentum.
4)Calculate total initial and final K.E of the system .
The difference is stored as P.E in the spring = 1/2ke^2.

Can you go from here ?

7. Jun 13, 2006

### Delzac

Yup i got the ans thx! :)

8. Jun 13, 2006

### arunbg

You're welcome :)