# Issues determining velocity in a totally elastic collision

A block of mass $M_1$ is attached by string to a support. The block is raised to a height and released. It then strikes a block of mass $M_2$ on a frictionless surface. Find the velocity of block $M_2$, assuming a totally elastic collision.

I have calculated the velocity ($v_1$) of $M_1$:
KE=PE
.5$M_1$$v_1^2=M_1gh v_1=\sqrt(2gh) From there I try to calculate the velocity (v'_2) of M_2: I use the conservation of momentum to solve for v'_2 M_1$$v_1$+$M_2$$v_2=M_1$$v'_1$+$M_2$$v'_2 and I assume that v_1 = \sqrt(2gh) v_2 = 0 v'_1 = 0 (can I assume this since I am not sure of the actual masses?) and then solve for v'_2 Doing this I get v'_2 = \frac{M_1}{M_2}\sqrt(2gh) , yet the answer given is v'_2 = \frac{2M_1}{M_1+M_2}\sqrt(2gh) If I don't assume v'_1 = 0 , then I get v'_2 = \frac{M_1}{M_2}(\sqrt(2gh)-v'_1) Can someone explain what I am doing wrong? I think there is something fundamentally that I am missing. Thanks for your time! • What does it mean "elastic"? How does you answer change if the collision isn't elastic? – Brian Moths Sep 10 '15 at 20:30 • Inelastic collisions involve both masses moving as a singular mass with a singular velocity after collision. M_1$$v_1$+$M_2$$v_2$=($M_1$+$M_2$)$v_3$ – slick1092 Sep 10 '15 at 20:40
• In an elastic collision, $KE = KE'$. No energy is lost in the collision. Since the masses are not stated to be equal, you must assume them different, so $v_1'$ cannot be assumed to be 0. – BowlOfRed Sep 10 '15 at 21:05