How far does the putty-block system compress the spring?

[eagles12]

Homework Statement

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

Homework Equations

E1=E2

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

 

[BruceW]

1/2 (.052+.460)(.203^2)=1/2(22)x^2
i got x=.042 but it is saying this is incorrect

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

 

[emailanmol]

Your equation is correct.
Plug in your values again and find x again.

 

[azizlwl]

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

 

[emailanmol]

M2 has no initial energy.
Total energy of the system only 1/2 (.052)(.203^2)
Final energy when all converted to potential energy is 1/2(22)x^2

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 :-)

 

[emailanmol]

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

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