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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.
block m_1 on string raised to height h
I have calculated the velocity ($v_1$) of $M_1$:
KE=PE
.5$M_1$$v_1^2$=$M_1$gh
$v_1$=$\sqrt(2gh)$

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!

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  • $\begingroup$ What does it mean "elastic"? How does you answer change if the collision isn't elastic? $\endgroup$ – Brian Moths Sep 10 '15 at 20:30
  • $\begingroup$ 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$ $\endgroup$ – slick1092 Sep 10 '15 at 20:40
  • $\begingroup$ 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. $\endgroup$ – BowlOfRed Sep 10 '15 at 21:05

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