A 1.0-kg ball of putty is released from rest and falls vertically 1.5 m until it strikes a hard floor, where it comes to rest in a 0.045-s time interval. what is the magnitude and direction of the average force exerted on the ball by the floor during the collision

Question

Physics

Emilie Booth

27 July, 23:55

A 1.0-kg ball of putty is released from rest and falls vertically 1.5 m until it strikes a hard floor, where it comes to rest in a 0.045-s time interval. what is the magnitude and direction of the average force exerted on the ball by the floor during the collision

+2

Answers (1)

Know the Answer?

Derrick Shields · Answer 1

We first calculate the velocity of the ball when it hits the ground; this is equal to the square root of the quantity (2*g*d) where g is the acceleration of gravity (9.8 m/s^2) and d is the distance fallen, 1.5m.

So, we get a velocity of sqrt (2*9.8*1.5) = 5.42 m/s.

We can calculate the impulse force applied to the putty ball by using Newton's second law, which states that the applied force is equal to the product of mass and acceleration, where acceleration can be further decomposed as the change in velocity divided by the change in time. Thus, inputting the known values, we have:

F = ma = m (dv/dt) = 1.0*5.42/0.045 = 120.4 newtons.