A billiard ball, labeled $1,$ moving horizontally strikes another billiard ball, labled $2,$ at rest. Before the impact ball $1$ was moving at a speed of $5\mathrm{.}00$ m$/$s and after the impact it is moving at a speed of $1\mathrm{.}00$ m$/$s at an$$angle of $50\mathrm{.}0$ degrees from the original direction. The two balls have equal mass.$$

$($b$)$Write conservation of momentum equations for the x and y directions separately and solve for$$the speed$$of ball $2$ after the impact.

$($d$)$Calculate the magnitude of the force resulting from the change in the player's velocity during the collision:

$($e$)$ Calculate the force due to the weight of the player. The forces of friction are usually given by$$xFk$=$kN$=$kmg$$are a fraction of the weight $($ie$.$ proportional to the weight$).$ Make the approximation that the frictional force$$is$$equal to the player's weight.

$($f$)$ Compare the two forces $($in this part, use xmg to approximate xFk$)$: