# Block hits block with spring. Conserv of Energy and Momentum?

1. Apr 6, 2005

### Fanman22

Below is a copy of the problem and questions that are asked. I'm pretty sure that this has to be a conservation of energy problem, but I keep getting the wrong answer for the first problem so I can't continue and try the others.
I set the KE of the moving block equal to the PE of the spring that it strikes to determine X meters. I get 0.41m....which is wrong. Any help is greatly appreciated.

A 3.0 kg block slides along a frictionless tabletop at 7.0 m/s toward a second block (at rest) of mass 6.0 kg. A coil spring, which obeys Hooke's law and has spring constant k = 860 N/m, is attached to the second block in such a way that it will be compressed when struck by the moving block, Fig. 9-40.

What will be the maximum compression of the spring?
m
(b) What will be the final velocities of the blocks after the collision? (Assume the initial direction of the 3.0 kg block is positive.)
3.0 kg block
m/s
6.0 kg block
m/s

(c) Is the collision elastic? (Compare energy values to the nearest joule.)

yes***
no

Give the kinetic energy of the system before and after the collision to justify your answer.
before
J
after
J

2. Apr 6, 2005

### dextercioby

The way i see it,linear momentum is conserved in the collision,but the total mechanical energy is conserved as well in 2 situations:globally and in the system spring-mass...

$m_{1}$ projectile's mass, $m_{2}$ target mass...

$$m_{1}\frac{\vec{v}_{1}^{2}}{2}=m_{1}\frac{\vec{v'}_{1}^{2}}{2}+m_{2}\frac{\vec{v'}_{2,MAX}^{2}}{2}$$

$$m_{1}\vec{v}_{1}=m_{1}\vec{v'}_{1}+m_{2}\vec{v'}_{2,MAX}$$

and of course

$$\frac{kA^{2}}{2}=\frac{m_{2}\vec{v'}_{2,MAX}^{2}}{2}$$

So i'd say that in the collision,the total KE is conserved,but after the bodies would no longer interact,in the next bilionth of the second,the velocity of tha target drops below the value aquired as a result of the collision,so the total balance is not valid anymore,till the body gets the same KE again,which is after T/2 and so on,because the target would oscillate forever with $T=2\pi \sqrt{\frac{m_{2}}{k}}$

So the last question is vague...

Daniel.

3. Apr 6, 2005

### Fanman22

ok, basically you lost me after you said that linear momentum is conserved....but if that is the case, shouldn't the initial KE of the block be equal to the PE of the compressed spring to give me the distance that it compressed?

4. Apr 6, 2005

### dextercioby

It would be the case,if the projectile had transferred all its KE in the collision.

U have to solve the system implied by the KE & momentum conservation in the collision

Daniel.

5. Apr 6, 2005

### Fanman22

my equation is as follows:
.5(3)(7)^2 = .5(860)x^2
x=0.41m
...but its wrong.

This is strange...if the surface is frictionless, then should the spring even compress at all? Wouldn't ANY amount of force be directly transfered into velocity since there is no friction holding the second block in place?

6. Apr 6, 2005

### Staff: Mentor

It's wrong because you are setting the total energy equal to the spring potential energy. But even when the spring is maximally compressed, the system is moving so it has KE as well.

Hint: When the spring is at maximum compression, how does the speed of each mass compare to the speed of the center of mass of the system? (What's the speed of the center of mass?)

The spring is attached to a mass, which has inertia.

7. Apr 6, 2005

### Fanman22

thanks, you helped me figure it out.