Impulse from position time graph

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

I think none of the options are correct . It is a uniform motion between t=0 and t=2 sec. Velocity is constant .

Impulse is given by m∆v .Since ∆v = 0 , impulse is zero at t= 2 sec .

Is that correct ?

 

[stockzahn]

Impulse is given by m∆v .Since ∆v = 0 , impulse is zero at t= 2 sec .

Is that correct ?

Unfortunately not. The impulse is the product of mass and velocity. $m\Delta v$ is the change in momentum $\Delta p$, from which you can calculate the acting force, if you know the time of the velocity change ($F=m\frac{dv}{dt}$).

 

[Jahnavi]

The impulse is the product of mass and velocity. $m\Delta v$ is the change in momentum $\Delta p$

Sorry . This is conceptually wrong .

Product of mass and velocity is momentum . Impulse is change in momentum .

 

[Orodruin]

It is a uniform motion between t=0 and t=2 sec. Velocity is constant .

This is not the question. The question is what the impulse given at t = 2 s is.

 

[Orodruin]

Sorry . This is conceptually wrong .

No it is not.

Product of mass and velocity is momentum . Impulse is change in momentum .

Which means that the change in momentum (with constant mass) is the mass multiplied by the change in velocity, i.e., $\Delta p = m\, \Delta v$, just as stated in #2.

 

[Jahnavi]

This is not the question. The question is what the impulse given at t = 2 s is.

You are right . Thanks !

 

[stockzahn]

Sorry . This is conceptually wrong .

Product of mass and velocity is momentum . Impulse is change in momentum .

Yes you are right, I'm sorry - language problem.

 

[Jahnavi]

No it is not

It is .

Please read post#2 again . I objected to the definition of impulse .

 

[Jahnavi]

Yes you are right, I'm sorry - language problem.

No problem