Aball of clay with a mass of 1.93kg hits the floor and comes to a stop. just
before it hit the floor, its velocity was -15.31m/s. if the collision took 0.33s, how
much force did the floor exert on the ball of clay?

89·539 N

Explanation:

Given the mass of the ball = 1·93 kg

Assuming that motion in upward direction is taken as positive

Velocity just before collision = - 15·31 m/s

As velocity is negative, it means that velocity is in downward direction as we have assumed that motion in upward direction as positive

Let the F be the force that the floor exert on the ball of clay

F = dp ÷ dt

where

dp is a small change in momentum

dt is a small change in time taken to cause dp

Here we can neglect force of gravity because the force that the floor will exert will be an impulsive force which means it is a large force which acts in a small interval of time

As this is a large force, the force of gravity will be negligible when compared to this large force

In this case

dp = d(m × v) = m × dv (∵ m is constant)

dv = final velocity - initial velocity = 0 - (-15·31) = 15·31 m/s

dp = 1·93 × 15·31 = 29·548 kg m/s

dt = 0·33 s

∴ F = 29·548 ÷ 0·33 = 89·539 kg m/s² = 89·539 N

As the force is positive, it means that force will act in upward direction

∴ The force that the floor exert on the clay = 89·539 N

the right answer is 0.3 meters per second