Simple change in momentum problem

[hms.tech]

Homework Statement

What is the rate of change of momentum when a constant 4N force acts on a mass of 2Kg for 2 seconds ?

Homework Equations

change in momentum = impulse produced (on the 2Kg mass) by the force
ie
f*t=δP

The Attempt at a Solution

i used the above equation to first find the impule
I,impulse= 4*2=8Ns
as I = δP
hence δP/δt = 8/2 = 4 Kg m s^-2

In simple terms as force = rate of change of momentum thus 4N shud be the answer...

but my teacher rejected the answer 4 but instead said that it would be 8

 

[TaxOnFear]

The rate of change of momentum is written in Newton's second law such that:

F = (Δmv)/Δt

You have m, you have t and you have F.

However, the rate of change of momentum doesn't include a force in it's equation, just like momentum = mv, change in momentum = (Δmv)/Δt.

I also get 4kgm/s² (or Newtons).

 

[hms.tech]

yeah, i think probably the teacher made a mistake, but i still need confirmation from other people as well ... so anyone else who could reconfirm this ?

 

[physicsvalk]

The rate of change of momentum is written in Newton's second law such that:

F = (Δmv)/Δt

You have m, you have t and you have F.

However, the rate of change of momentum doesn't include a force in it's equation, just like momentum = mv, change in momentum = (Δmv)/Δt.

I also get 4kgm/s² (or Newtons).

The time derivative of momentum is force (assuming we're talking about a linear system, and we're not talking relativistically). It can include force if a net force is causing it to change.

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Impulse is the change in momentum of a system, and is equal to the force acting on it. You're getting confused with the problem. Once you solve for the impulse, you've found the object's momentum (assuming that P_0= 0). It looks like, to me, that after you found the impulse, you divided by the time interval. This would in turn give you the force that was exerted on the object. Your teacher would be correct if the question was asking for the final momentum on the system.

Another way to check is to look at the units of your answer.