Putty on Block and Spring

1. Mar 22, 2012

eagles12

1. The problem statement, all variables and given/known data

a .46kg block is attached to a horizontal spring that is at equilibrium length, and whose force constant is 22 N/m. The block rests on a frictionless surface. A 5.2x10-2 wad of putty is thrown horizontally at the block, hitting it with a speed of 2 m/s and sticking.

How far does the putty-block system compress the spring

2. Relevant equations

E1=E2

3. The attempt at a solution

K1U1=K2U2
1/2 (m1+m2) v^2=1/2Kx^2
1/2 (.052+.460)(.203^2)=1/2(22)x^2
i got x=.042 but it is saying this is incorrect

2. Mar 22, 2012

BruceW

The equation looks good, but I don't get x=.042, maybe check the calculation again.

3. Mar 22, 2012

emailanmol

Plug in your values again and find x again.

4. Mar 22, 2012

azizlwl

Yes i was wrong. I always confuse between conservation of energy and conservation of momentum.

Last edited: Mar 23, 2012
5. Mar 23, 2012

emailanmol

Hey Azizlwl,

Actually OP's equation is correct.

You cannot apply conservation of energy from start because its an inelastic collision.
Therefore, energy before colllision is not equal to energy after collision.
For velocities after collision, he has used conservation of momentum and thus derived the new energy.

Energy is conserved during the course of journey after the collision, which is exactly the equation OP used :-)

6. Mar 23, 2012

emailanmol

No problem at all.Happens to most of us :-)