Law of conservation of linear momentum:

$\overline{){\mathbf{\sum}}{{\mathbf{P}}}_{{\mathbf{i}}}{\mathbf{=}}{\mathbf{\sum}}{{\mathbf{P}}}_{{\mathbf{f}}}}$

**Part A**

From the law of conservation of linear momentum, we have:

$\begin{array}{rcl}{\mathbf{m}}_{\mathbf{1}}{\mathbf{u}}_{\mathbf{i}}\mathbf{+}\mathbf{0}& \mathbf{=}& {\mathbf{m}}_{\mathbf{1}}{\mathbf{u}}_{\mathbf{f}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}{\mathbf{v}}_{\mathbf{f}}\\ {\mathbf{m}}_{\mathbf{1}}\mathbf{(}{\mathbf{u}}_{\mathbf{i}}\mathbf{-}{\mathbf{u}}_{\mathbf{f}}\mathbf{)}& \mathbf{=}& {\mathbf{m}}_{\mathbf{2}}{\mathbf{v}}_{\mathbf{f}}\end{array}$

Block 1, of mass m_{1}, moves across a frictionless surface with speed u_{i}. It collides elastically with block 2, of mass m_{2}, which is at rest (v_{i} = 0). (Figure 1) After the collision, block 1 moves with speed u_{f}, while block 2 moves with speed v_{f}. Assume that m1 > m2, so that after the collision, the two objects move off in the direction of the first object before the collision.

Part A

What is the final speed *u*_{f} of block 1? Express *u*_{f} in terms of *m*_{1}, m_{2}, and *u*_{i}.

Part B

What is the final speed *v*_{f} of block 2? Express *v*_{f} in terms of *m*_{1}, m_{2}, and *u*_{i}.

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