Law of conservation of energy:

$\overline{)\begin{array}{rcl}\mathbf{P}\mathbf{.}{\mathbf{E}}_{\mathbf{i}}\mathbf{+}\mathbf{K}\mathbf{.}{\mathbf{E}}_{\mathbf{i}}& {\mathbf{=}}& \mathbf{P}\mathbf{.}{\mathbf{E}}_{\mathbf{f}}\mathbf{+}\mathbf{K}\mathbf{.}{\mathbf{E}}_{\mathbf{f}}\\ \mathbf{m}\mathbf{g}{\mathbf{h}}_{\mathbf{i}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{m}{{\mathbf{v}}_{\mathbf{i}}}^{\mathbf{2}}& {\mathbf{=}}& \mathbf{m}\mathbf{g}{\mathbf{h}}_{\mathbf{f}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{m}{{\mathbf{v}}_{\mathbf{f}}}^{\mathbf{2}}\end{array}}$

Momentum:

$\overline{){\mathbf{p}}{\mathbf{=}}{\mathbf{m}}{\mathbf{v}}}$

Impulse:

$\overline{){\mathbf{J}}{\mathbf{=}}{\mathbf{\u2206}}{\mathbf{p}}}$

or

$\overline{){\mathbf{J}}{\mathbf{=}}{\mathbf{F}}{\mathbf{t}}}$

**a.**

From law of conservation of energy,

mgh_{i} + 0 = 0 + (1/2)mv_{i}^{2}

gh_{i} = (1/2)v_{i}^{2}

v_{i} = sqrt (2gh_{i}) = sqrt[(2)(9.8)(1.5)] = 5.42 m/s

As shown in the figure (Figure 1) , a superball with mass *m* equal to 50 grams is dropped from a height of *h*_{i} =1.5 m. It collides with a table, then bounces up to a height of *h*_{f} = 1.0 m. The duration of the collision (the time during which the superball is in contact with the table) is *t*_{c} = 15ms. In this problem, take the positive *y* direction to be upward, and use *g *= 9.8 m/s^{2} for the magnitude of the acceleration due to gravity. Neglect air resistance.

a. Find the *y* component of the momentum, *p*_{before,}* _{y}*, of the ball immediately before the collision.

b. Find the *y* component of the momentum of the ball immediately after the collision, that is, just as it is leaving the table.

c. Find *J** _{y}*, the

d. Find the *y* component of the time-averaged force *F*_{avg,}* _{y}*, in newtons, that the table exerts on the ball.

e. Find *K*_{after} - *K*_{before}, the change in the kinetic energy of the ball during the collision, in joules.

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