# Combining Conservation Laws

1. Nov 7, 2006

### merlos

Having a difficult time determining which laws and equations to use?

A 15.0 kg block is attached to a very light horizontal spring of force constant 350 N/m and is resting on a smooth horizontal table. Suddenly it is struck by a 3.00 kg stone traveling horizontally at 8.00m/s to the right, whereupon the stone rebounds at 2.00 m/s horizontally to the left.

Find the maximum distance in meters that the block will compress the spring after the collision.

2. Nov 7, 2006

### Staff: Mentor

Think of it as two problems. Take them one at a time:
(1) The collision of stone with block--what's conserved there?
(2) The compression of the spring--what's conserved here?

3. Nov 7, 2006

### merlos

Combining Conservation Laws; where'd I go wrong?

A 15.0 kg block is attached to a very light horizontal spring of force constant 350 N/m and is resting on a smooth horizontal table. Suddenly it is struck by a 3.00 kg stone traveling horizontally at 8.00 m/s to the right, whereupon the stone rebounds at 2.00 m/s horizontally to the left.

Don't know where I went wrong?
Here's my setup:

Collision:

Change in KE(stone) = (1/2)(m)(vfsquared-visquared)
= (1/2)(3.00kg)(-2m/s squared - 8 m/s squared)
= -90

After the collision:

Change in KE(stone) = (1/2)(Force constant)xsquared
-90 = (1/2)(350N/m)xsquared
x = 0.72m

4. Nov 7, 2006

### Staff: Mentor

Start by answering the questions I posed in my earlier response.

5. Nov 7, 2006

### merlos

1.) Kinetic energy
2.) Kinetic energy of block is transferred to the spring, where it's stored as potential energy

6. Nov 7, 2006

### Staff: Mentor

In general, KE is not conserved in a collision. But what is conserved in every collision?

Right!

7. Nov 7, 2006

1.) Momentum

8. Nov 7, 2006

### Staff: Mentor

Right. Now use that to find the speed of the block just after the collision.