Force and change in momentum with time

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A constant force of 6.00 N acting on a 3.00-kg object for 10.0 seconds results in a change in momentum that can be calculated using the impulse-momentum theorem, where impulse equals the change in momentum. The first problem involves calculating the change in velocity, which can be derived from the equation F = (delta p) / (delta t). For the second problem, the change in momentum of the 625-kg car can be found using the formula p = m(v_final - v_initial), leading to a need for clarity on whether to include delta in the calculations. The discussion emphasizes the importance of understanding the relationship between force, momentum, and time, highlighting that F = dp/dt is universally applicable. Ultimately, the key takeaway is that impulse equals the change in momentum, reinforcing fundamental physics concepts.
chowyo123
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Hey, It's been about 2 weeks since I've had a physics class and I just can't seem to put things together again.
I have two problems that seem like they should be easy, and I'm sure they are, but am unsure on how to do them.

A constant force of 6.00N acts on a 3.00-kg object for 10.0 s. What are the changes in the objects momentum and velocity?

I have the equation momentum p = mv. p=(F)(delta t). Could I just do 6.00-N x 10=(3.00-kg)v and solve for v to find the velocity?

The second problem is
The velocity of a 625-kg car is changed from 10.0 m/s to 44.0 m/s in 68.0 s by an external constant force. What is the resulting change in momentum of the car? What is the magnitude of the force?
I tried to do p=(625)(44-10.0) to find the momentum, but that does not seem right to me. What do i do?
 
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For question one I don't think your looking for a velocity, your're looking for a change in velocity.
 
(delta p) = F (delta t) ... there has to be a change on both sides !

or only one side, with a ratio on the other side : F = (delta p) / (delta t).

or nowhere, as in : p = m v .

now, in your (625)(44 - 10) , does the other side have a delta, or not?

and whatever happened to UNITS ?!
 
The impulse that an object experiences is equal to the change in it's momentum!
 
Alright, I ended up remembering how to do these problems.
 
Yeah, just remember that F=ma only works at low velocities, but F=\frac{dp}{dt} is true always.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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