Posted by **Carden** on Thursday, April 29, 2010 at 8:27pm.

A 0.409 kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 19.5 N/m. The block rests on a frictionless surface. A 0.0500 kg wad of putty is thrown horizontally at the block, hitting it with an initial speed of 2.32 m/s and sticking. How far does the putty-block system compress the spring?

My answer .366 is 10% off. Why/How?

- Physics -
**bobpursley**, Thursday, April 29, 2010 at 8:30pm
How can I possibly know why your answer is off? I have no idea what you did.

- Physics -
**Carden**, Thursday, April 29, 2010 at 8:33pm
I did m1v1 /(m1+m2) = vf = .252723

x = vf(m/k)

x = .366

- Physics -
**bobpursley**, Thursday, April 29, 2010 at 8:35pm
after getting vf, you need to calculate the KE of the putty/block, and set that equal to the compressed energy in the spring 1/2 k x^2, and solve for x.

I think you need a tutor, you appear to be totally lost.

- Physics -
**Carden**, Thursday, April 29, 2010 at 8:36pm
x = v * sqrt(m/k)

correct?

- Physics -
**bobpursley**, Thursday, April 29, 2010 at 8:39pm
No.

Set the KE of the putty/block, set it equal to the max stored energy of the spring, 1/2 k x^2

- Physics -
**Carden**, Thursday, April 29, 2010 at 8:40pm
when I set 1/2mv2 equal to 1/2kx2

I got .366

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