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PHY 231 HW9 More energy and momentum View Basic/Answers
HW9 More energy and momentum Begin Date: 3/23/2015 12:00:00 PM ­­ Due Date: 4/1/2015 11:59:00 PM End Date: 4/1/2015 11:59:00 PM Problem 1: Problem 2: Problem 3: Problem 4: Problem 5: Problem 6: Problem 7: Problem 8: Problem 9: Problem 10: Problem 11: Problem 12: Full solution not currently available at this time.
Rafael drops a hard rubber ball, of mass m = 105 g, onto the pavement. He
drops it from a height h1 = 2.5 m, and it bounces back to a height h2 = 1.5 m.
While the ball is in contact with the ground, the ground exerts a non­
constant force on the ball as shown in the figure. During the time interval
Δt1, the force rises linearly to Fmax, and during the time interval Δt2, the
force returns to zero as the ball leaves the ground. Randomized Variables
m = 105 g
h1 = 2.5 m
h2 = 1.5 m Variable Name
Min
Max
Step
Sample Value
m
100
325
5
105
h1
2.5
3.5
0.1
2.5
h2
1.5
2.35
0.05
1.5
t1
10
20
1
11
t2
40
80
5
45
Part (a) What is the magnitude of the impulse delivered to the ball by the ground, in terms of Fmax and the time
intervals Δt1 and Δt2?
Correct Equation: I = 0.5 Fmax ( Δt1 + Δt2 )
Choice Info:
Hints:
1 hints available
Valid Choices:
0.5, Δt1, Fmax, 0.5, Δt2, Fmax, The impulse given to the ball is the integral of
the force over the time interval ­ but this is also
just the area under the curve shown in the figure.
Partial Credit Choices with Feedback:
InValid Choices:
t, α, m, θ, g, d, β, h, P, j, k, a, Part (b) If the time intervals are Δt1 = 11 ms and Δt2 = 45 ms, what is the magnitude of the maximum force between
the ground and the ball, in newtons?
Correct Algorithm: Fmax = 2*m/1000*((2*9.81*h2)^0.5+(2*9.81*h1)^0.5)/(t1/1000+t2/1000)
Choice Info:
Fmax = 2*105/1000*((2*9.81*1.5)^0.5+
(2*9.81*2.5)^0.5)/(11/1000+45/1000)
Fmax = 2*105/1000*((2*9.81*1.5)^0.5+
(2*9.81*2.5)^0.5)/(11/1000+45/1000)
Fmax = 46.607
Buffer + or ­ 1.39821
Problem 13: Hints:
2 hints available
You know how the impulse is related to the
maximum force from part (a), but how can you
find another way to calculate the impulse? Do
you know the momentum before and after the
collision with the ground?
The velocity before and after the collision can be
found using conservation of energy, which will
give you the change in momentum during the
collision with the ground.
Problem 14: Problem 15: Problem 16: All content © 2014 Expert TA, LLC