Conservation of Mechanical Energy in a ball of clay

[octahedron]

Homework Statement

A 50 g ball of clay traveling at speed v_0 hits and sticks to a 1.0 kg block sitting at rest on a frictionless surface.

a. What is the speed of the block after the collision?
b. Show that the mechanical energy is not conserved in this collision. What percentage of the ball's initial energy is "lost?"

The Attempt at a Solution

For part (a), I just said

m_1*v_0 = (m_1 + m_2)*v_1

0.05*v_0 = (0.05+1)*v_1

v_0 = 21*v_1

I'm hoping this is correct! Now, part (b) I'm not entirely sure of. I guess there is no PE in this case so I should show that

(.5)(m_1)(v_0)^2 \neq (.5)(m_1)(v_0)^2 + (.5)(m_1)(v_1)^2

Is this it? How do I go on about finding the percentage "lost"?

 

[Doc Al]

For part (a), I just said

m_1*v_0 = (m_1 + m_2)*v_1

0.05*v_0 = (0.05+1)*v_1

This is fine.

v_0 = 21*v_1

Not sure what happened here. Find v_1 in terms of v_0.

I'm hoping this is correct! Now, part (b) I'm not entirely sure of. I guess there is no PE in this case so I should show that

(.5)(m_1)(v_0)^2 \neq (.5)(m_1)(v_0)^2 + (.5)(m_1)(v_1)^2

This expression seems a bit mixed up. Find the initial KE (the left hand side). Find the final KE (you'll need the speed of the combined masses from part a). (Express everything in terms of v_0.)

Compare the two KEs and find their difference (the loss of KE). Then you can figure out what percentage of the original KE that loss is.

 

[Supernats]

Is the right side of the equation your after? If it is, v_0 shouldn't figure into it, I don't believe.

 

[rl.bhat]

Initial KE = Ei = 0.5*m1*Vo^2
Final KE = Ef = 0.5*(m1 + m2 )*V1^2
Show that Ef is not equal to Ei. And percentage loss = [(Ei -Ef)/Ei]*100

 

[octahedron]

Thanks everyone, got it! :)

Is the right side of the equation your after? If it is, v_0 shouldn't figure into it, I don't believe.

Oops! I didn't mean to include v_0 there. Sorry for the confusion.