<p>Nevermind - got it.</p>
<p>edit - ok i guess you got it lol</p>
<p>Thanks anyway. I have another question from the CB book.</p>
<p>(A) Conservation of energy alone
...
(C) Conservation of both energy and momentum
...</p>
<ol>
<li>Used to calculate the speed of a lump of clay that hits and sticks to a block of wood suspended as a pendulum, given the height to which the block swings and the masses of the block and the clay</li>
</ol>
<p>The correct answer is C, but I put A.</p>
<p>Isn't using KE lost = GPE gained sufficient?</p>
<p>If the mass of clay = m, mass of block = M, speed of clay = v, height = h, then does 0.5mv^2 = (m+M)gh?</p>
<p>you are right
0.5mv^2= (m+M)gh is true, but momentum is conserved as well ( there is a collision between the block and the clay )
so you will have : mClay*vi = (M+m)vf because i suppose that the wooden block is stationary before the impact isn't it ?</p>