A steel ball with mass 36.0 gg is dropped from a height of 2.01 mm onto a horizontal steel slab. The ball rebounds to a height of 1.67 mm.Part ACalculate the impulse delivered to the ball during impact.Express your answer in newton seconds.Activate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value typenothingN⋅sN⋅sSubmitRequest AnswerPart BIf the ball is in contact with the slab for a time of 1.60 msms, find the average force on the ball during impact.Express your answer in newtons.
Question
A steel ball with mass 36.0 gg is dropped from a height of 2.01 mm onto a horizontal steel slab. The ball rebounds to a height of 1.67 mm.Part ACalculate the impulse delivered to the ball during impact.Express your answer in newton seconds.Activate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value typenothingN⋅sN⋅sSubmitRequest AnswerPart BIf the ball is in contact with the slab for a time of 1.60 msms, find the average force on the ball during impact.Express your answer in newtons.
Solution
Part A
First, we need to convert the given values to the correct units. The mass of the ball is given in grams, but we need it in kilograms for our calculations. Similarly, the heights are given in millimeters, but we need them in meters.
So,
m = 36.0 g = 0.036 kg h1 = 2.01 mm = 0.00201 m h2 = 1.67 mm = 0.00167 m
We know that the initial velocity (v1) of the ball before it hits the slab is obtained by using the equation of motion:
v1 = sqrt(2gh1)
where g is the acceleration due to gravity (9.8 m/s²).
After the ball rebounds, it has a final velocity (v2) which can be obtained by:
v2 = sqrt(2gh2)
The change in velocity (Δv) of the ball is then:
Δv = v2 - (-v1) = v2 + v1
The impulse delivered to the ball during impact is the product of the mass of the ball and the change in velocity, which is:
Impulse = m * Δv
Part B
The average force on the ball during impact is the impulse divided by the time of contact. The time is given in milliseconds, but we need it in seconds for our calculations.
t = 1.60 ms = 0.00160 s
So,
Average Force = Impulse / t
Now you can substitute the values into the equations and calculate the answers.
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