The conservation of linear momentum equation:

$\overline{)\begin{array}{rcl}{\mathbf{m}}_{\mathbf{1}}{\mathbf{v}}_{\mathbf{01}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}{\mathbf{v}}_{\mathbf{02}}& {\mathbf{=}}& {\mathbf{m}}_{\mathbf{1}}{\mathbf{v}}_{\mathbf{f}\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}{\mathbf{v}}_{\mathbf{f}\mathbf{2}}\end{array}}$

We'll also use the law of conservation of energy:

$\overline{){{\mathbf{K}}}_{{\mathbf{i}}}{\mathbf{+}}{{\mathbf{U}}}_{{\mathbf{i}}}{\mathbf{+}}{{\mathbf{W}}}_{{\mathbf{nc}}}{\mathbf{=}}{{\mathbf{K}}}_{{\mathbf{f}}}{\mathbf{+}}{{\mathbf{U}}}_{{\mathbf{f}}}}$, where W_{nc} is the work done by non-conservative forces such as friction.

For our case, U_{i} = 0 since there is no height (Remember U = mgh).

W_{nc} = 0 because there is no friction or any other non-conservative forces (we're told that the spring is on a smooth horizontal table).

After the collision, we do not have kinetic energy since the spring will be at rest.

A 15.0 kg block is attached to a very light horizontal spring of force constant 475 N/m and is resting on a smooth horizontal table. (See the figure below .) Suddenly it is struck by a 3.00 kg stone traveling horizontally at 8.00 m/s to the right, whereupon the stone rebounds at 2.00 m/s horizontally to the left.

Find the maximum distance that the block will compress the spring after the collision.(*Hint*: Break this problem into two parts - the collision and the behavior after the collision - and apply the appropriate conservation law to each part.)

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