How Do Conservation Laws Apply to a Collision and Subsequent Motion?

AI Thread Summary
The discussion focuses on applying conservation laws to analyze a collision between a stone and a block attached to a spring. Key points include the conservation of momentum during the collision, which is crucial for determining the block's speed post-collision. The kinetic energy of the block is converted into potential energy in the spring as it compresses. The calculations provided indicate a misunderstanding of energy conservation, particularly regarding kinetic energy not being conserved in collisions. The final goal is to find the maximum compression of the spring after the collision, emphasizing the importance of correctly applying conservation principles.
merlos
Messages
14
Reaction score
0
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.
 
Physics news on Phys.org
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?
 
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
 
Please don't start a second thread on the same problem.

Start by answering the questions I posed in my earlier response.
 
1.) Kinetic energy
2.) Kinetic energy of block is transferred to the spring, where it's stored as potential energy
 
merlos said:
1.) Kinetic energy
In general, KE is not conserved in a collision. But what is conserved in every collision?

2.) Kinetic energy of block is transferred to the spring, where it's stored as potential energy
Right!
 
1.) Momentum
 
Right. Now use that to find the speed of the block just after the collision.
 

Similar threads

Back
Top