Momentum and Impulse: Dropping an Object

[Judah]

1. A 2 kilogram egg drop apparatus began at rest and fell through approximately 5-m. Assume that it’s momentum was changed to zero in a time of 0.25-s.
--What force stopped your apparatus?
---Mass=2kg
---Distance=-5m
---Time=.25-s

2.-Force(Time)=Mass(Velocity)

3.Velocity final for the apparatus would be the square root of 2(-9.8)(-5). About 9.99m/s.
Do i just plug in that for the equation above?

 

[ehild]

2.-Force(Time)=Mass(Velocity)

3.Velocity final for the apparatus would be the square root of 2(-9.8)(-5). About 9.99m/s.
Do i just plug in that for the equation above?

Equation 2 should be written as Force * time = Mass *(change of velocity)
Yes, you nee only plug in the data to get the average force of impact.

ehild

 

[Judah]

Thanks I got the answer for that. Can you answer another question?
If the object were to bounce would it require a larger, smaller, or the same force?

 

[ehild]

What is the change of the velocity if the object bounces?

ehild

 

[asteriatic]

Yes, you can plug in the final velocity of 9.99 m/s into the equation Force(Time) = Mass(Velocity) to calculate the force that stopped the apparatus. The force can be calculated by multiplying the mass of 2 kg with the velocity of 9.99 m/s, which gives a force of 19.98 N. This force is the impulse applied to the apparatus, which caused it to change its momentum to zero in 0.25 seconds. Therefore, the force that stopped the apparatus was 19.98 N.