Final answer:
The average force exerted by the eight ball on the cue ball is -153 N during a collision, implying a force of 153 N in the direction opposite to the cue ball's original motion.
Step-by-step explanation:
The student is asking to calculate the average force exerted by the eight ball on the cue ball during a collision in a game of pool. The cue ball has a mass of 0.17 kg, initially moves at 5.0 m/s, and then is knocked back at 4.0 m/s. The time of contact during the collision is 0.010 s.
To find the average force, we use the formula F = \(\frac{\Delta p}{\Delta t}\), where \(\Delta p\) is the change in momentum and \(\Delta t\) is the time over which the change occurs. The change in momentum \(\Delta p\) can be calculated as \(\Delta p = m(v_f - v_i)\), where m is the mass, \(v_i\) is the initial velocity, and \(v_f\) is the final velocity. Substituting the values we have,
\(\Delta p = 0.17 \, kg \, ( -4.0 \, m/s - 5.0 \, m/s ) = -1.53 \, kg\cdot m/s\).
Now, we can calculate the average force,
F = \(\frac{-1.53 \, kg\cdot m/s}{0.010 \, s} = -153 \, N\).
The average force is thus -153 N, which means the eight ball exerts a force of 153 N on the cue ball in the direction opposite to the original motion of the cue ball (since force is a vector and the negative sign indicates direction).