What is the net impulse and average force on a ball caught in hand?

AI Thread Summary
The net impulse exerted on a ball of mass 1.5 kg traveling downward at 10 m/s and coming to rest in 0.25 seconds is calculated to be 15 kg m/s. The average force exerted by the hand on the ball must account for gravitational force, which adds to the total force needed to stop the ball. The gravitational force acting on the ball is 14.7 N (1.5 kg multiplied by 9.8 m/s²). The hand must exert a force greater than 60 N to counteract both the gravitational force and to bring the ball to rest. Understanding the relationship between impulse and gravitational force is crucial for accurate calculations.
snoopygal327
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Homework Statement


A ball of mass 1.5 kg is traveling downward at a speed of 10 m/s. It comes to rest in your hand in .25 s. What is the net impulse exerted on the ball? What is the average force exerted by the hand on the ball (Don't forget the gravitational force.)

Homework Equations


J = mvf-mvi
J = F\Deltat

The Attempt at a Solution


I figured out that the impulse is 15 kg m/s from the first equation. I thought that I had figured out the force exerted by the hand on the ball to be 60 N but now I'm not sure what it means by gravitational force. How does this affect the force exerted by the hand on the ball?
 
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snoopygal327 said:

Homework Statement


A ball of mass 1.5 kg is traveling downward at a speed of 10 m/s. It comes to rest in your hand in .25 s. What is the net impulse exerted on the ball? What is the average force exerted by the hand on the ball (Don't forget the gravitational force.)

Homework Equations


J = mvf-mvi
J = F\Deltat

The Attempt at a Solution


I figured out that the impulse is 15 kg m/s from the first equation. I thought that I had figured out the force exerted by the hand on the ball to be 60 N but now I'm not sure what it means by gravitational force. How does this affect the force exerted by the hand on the ball?

Over that time period that you used you also had gravity in the form of m*g acting on the hand as well.

For instance if you had merely arrested its acceleration at the point that it contacted your hand and it traveled still at 10m/s with your hand then your hand would have been exerting a force of m*g on it to keep it from accelerating further. Not only did you do that, but you also arrested its motion all together which is the impulse force you calculated.
 
LowlyPion said:
Over that time period that you used you also had gravity in the form of m*g acting on the hand as well.

For instance if you had merely arrested its acceleration at the point that it contacted your hand and it traveled still at 10m/s with your hand then your hand would have been exerting a force of m*g on it to keep it from accelerating further. Not only did you do that, but you also arrested its motion all together which is the impulse force you calculated.

I have to calculate another impulse for the gravitational force? So it would be something like J = (1.5 kg)(9.8 m/s2)(.25s).
 
snoopygal327 said:
I have to calculate another impulse for the gravitational force? So it would be something like J = (1.5 kg)(9.8 m/s2)(.25s).

No. It's just that the gravitational force was constant throughout the impulse and hence is added to the average force.
 
Oh I see. Thank you.
 
snoopygal327 said:
Oh I see. Thank you.

No problem.

Cheers.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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