a crane drops a .3 kg steel ball onto a steel plate. the ball's speeds just before impact and after are 4.5 m/s and 4.2 m/s respectively. If the ball is in contact with the plate for .03 s, what is the magnitude of the average force that the ball exerts on the plate during impact?

Question

a crane drops a .3 kg steel ball onto a steel plate. the ball's speeds just before impact and after are 4.5 m/s and 4.2 m/s respectively. If the ball is in contact with the plate for .03 s, what is the magnitude of the average force that the ball exerts on the plate during impact?

Answer 1

Presumably the steel ball bounced and changed directions. The momentum change is therefore 0.3 kg*(4.5 + 4.2 kg.s) = 2.61 kg m/s

Set that equal to the impulse (Force * time) and divide by the time interval to get the average force.

Answer 2

To find the magnitude of the average force that the ball exerts on the plate during impact, we can use Newton's second law of motion. According to Newton's second law, the force is equal to the rate of change of momentum.

The momentum of an object is given by the product of its mass and velocity. In this case, the momentum just before impact (p1) is equal to the mass of the ball (m) multiplied by its speed just before impact (v1), and the momentum just after impact (p2) is equal to the mass of the ball (m) multiplied by its speed just after impact (v2).

p1 = m * v1
p2 = m * v2

Now, the change in momentum (Δp) can be calculated as:

Δp = p2 - p1

The average force (F) exerted on the plate during impact can be obtained by dividing the change in momentum by the time interval during which the contact is made:

F = Δp / Δt

Given that the mass of the ball (m) is 0.3 kg, the speed just before impact (v1) is 4.5 m/s, the speed just after impact (v2) is 4.2 m/s, and the time interval (Δt) is 0.03 s, we can now calculate the magnitude of the average force (F).

Step 1: Calculate the momentum just before impact (p1).
p1 = m * v1 = 0.3 kg * 4.5 m/s = 1.35 kg·m/s

Step 2: Calculate the momentum just after impact (p2).
p2 = m * v2 = 0.3 kg * 4.2 m/s = 1.26 kg·m/s

Step 3: Calculate the change in momentum (Δp).
Δp = p2 - p1 = 1.26 kg·m/s - 1.35 kg·m/s = -0.09 kg·m/s

Step 4: Calculate the magnitude of the average force (F).
F = Δp / Δt = (-0.09 kg·m/s) / (0.03 s) = -3 N

The magnitude of the average force that the ball exerts on the plate during impact is 3 Newtons (N).

a crane drops a .3 kg steel ball onto a steel plate. the ball's speeds just before impact and after are 4.5 m/s and 4.2 m/s respectively. If the ball is in contact with the plate for .03 s, what is the magnitude of the average force that the ball exerts on the plate during impact?

Presumably the steel ball bounced and changed directions. The momentum change is therefore 0.3 kg*(4.5 + 4.2 kg.s) = 2.61 kg m/s

87

To find the magnitude of the average force that the ball exerts on the plate during impact, we can use Newton's second law of motion. According to Newton's second law, the force is equal to the rate of change of momentum.

87 N