Question

A 28 g steel ball bounces elastically on a steel plate, always returning to the same maximum height h = 27.5 m.

1)

With what speed does the ball leave the plate?

v_up =

m/s  

2)

What is the magnitude of the total change in momentum of the ball with each bounce?

|Dp_ball| =

kg m/s  

3)

What is the magnitude of the total momentum the ball transfers to the plate at each bounce?

kg m/s  

4)

What is the time interval between bounces?

Dt =

s  

5)

Averaged over a long time interval, what is the magnitude of the rate at which momentum is transferred to the plate?

|Dp/Dt| =

N  

6)

What is the magnitude of the average force exerted by the ball on the plate?

F_avg =

N  

7)

What is the weight of the ball?

W =

N

Answer 1

(1) Again assuming a terrestrial environment, the "v-up" can be found from
2*27.5 m*9.8 m/s^2 = (v-up)^2
v-up = 23.21 m/s

(2) (0.028 kg)23.21 m/s)*2 = 1.299 kg - m/s

(3) 1.299 kg-m/s

(4) The time for JUST the falling satisfies h = (1/2)gt^2,
so the time for the rising AND falling must satisfy
t = 2*sqrt(2h/g) = 4.73 seconds

(5) 1.299 kg-m/s divided by 4.73 seconds
= 0.274 kg m/s^2

(6) 0.274 N

(7) is the easiest, if we're to assume this happens in a terrestrial environment! W = (0.028 kg)(9.8 m/s^2) = 0.274 N

Answer 2

(1) Again assuming a terrestrial environment, the "v-up" can be found from
2*27.5 m*9.8 m/s^2 = (v-up)^2
v-up = 23.21 m/s

(2) (0.028 kg)(23.21 m/s)*2 = 15.08 kg - m/s

(3) 15.08 kg-m/s

(4) The time for JUST the falling satisfies h = (1/2)gt^2,
so the time for the rising AND falling must satisfy
t = 2*sqrt(2h/g) = 4.738 seconds

(5) 15.08 kg-m/s divided by 4.738 seconds
= 3.182 kg m/s^2

(6) 3.182 N

7) W = (0.028 kg)(9.8 m/s^2) = 0.2744 N