Question

A billiard ball strikes and rebounds from the cushion of a pool table perpendicularly. The mass of the ball is 0.38 kg The ball approaches the cushion with a velocity of +2.20 m/s and rebounds with a velocity of -1.70 m/s. The ball remains in contact with the cushion for a time of 3.40 x 10^-3 s. What is the average net force (magnitude and direction) exerted on the ball by the cushion?

Answer 1

Force is 432.94 N along the rebound direction of ball.

Step-by-step explanation:

Force is rate of change of momentum.

`\texttt{Force}=\frac{\texttt{Final momentum-Initial momentum}}{\texttt{Time}}`

Final momentum = 0.38 x -1.70 = -0.646 kgm/s

Initial momentum = 0.38 x 2.20 = 0.836 kgm/s

Change in momentum = -0.646 - 0.836 = -1.472 kgm/s

Time = 3.40 x 10⁻³ s

`\texttt{Force}=\frac{\texttt{Final momentum-Initial momentum}}{\texttt{Time}}=(-1.472)/(3.40* 10^(-3))\\\\\texttt{Force}=-432.94N`

Force is 432.94 N along the rebound direction of ball.