Write a realistic word problem for which this is the correct equation

[ChetBarkley]

Homework Statement

(0.10kg)(40m/s)-(0.10kg)(-30m/s)=1/2(1400N)(delta t)

Relevant Equations

Impulse = change in momentum = Force * delta t

So far for the word problem I have: A 100 g particle, traveling at 40 m/s, collides inelastically with another 100g particle traveling towards it at 30 m/s.

Now from the equation provided we need the question to ask us to find delta t, and that's simple enough but I'm not sure what that 1/2 is doing front of the right hand side. Does it mean that we only want half the impulse, or that the particles collide elastically and we want the impulse of just one of them?

 

[PeroK]

Homework Statement:: (0.10kg)(40m/s)-(0.10)(-30m/s)=1/2(1400N)(delta t)
Relevant Equations:: Impulse = change in momentum = Force * delta t

So far for the word problem I have: A 100 g particle, traveling at 40 m/s, collides inelastically with another 100g particle traveling towards it at 30 m/s.

Why two particles? Why not one particle being hit with something? Or, colliding with something.

I'm not sure what that 1/2 is doing front of the right hand side.

Good question.

 

[ChetBarkley]

Why two particles? Why not one particle being hit with something? Or, colliding with something.

Good question.

Well if it were one particle hitting something, like a wall, then what would the 0.100kg and the -30m/s represent?

 

[kuruman]

Homework Statement:: (0.10kg)(40m/s)-(0.10)(-30m/s)=1/2(1400N)(delta t)

Shouldn't you have (0.10kg)(40m/s)-(0.10kg)(-30m/s)=1/2(1400N)(delta t)?

Also, with this correction, you cannot have two 0.10 kg particles. The right hand side is an impulse which means that the left hand side must be the inelastic momentum change of a single 0.10 kg particle. I don't understand the "1/2" but if it belongs there, I suspect it is the tricky part to this question.

 

[PeroK]

Well if it were one particle hitting something, like a wall, then what would the 0.100kg and the -30m/s represent?

Mass of particle and initial velocity?

 

[kuruman]

I don't understand the "1/2" but if it belongs there, I suspect it is the tricky part to this question.

I can now see how to get the 1/2 factor in. The impulse equation is $J=m\Delta v=\bar F \Delta t$. One has to cook up a force function $F(t)$ to describe a contact force parametrized by $F_0$ such that $$\bar F\equiv \frac{\int_0^{T} F(t)~dt}{\int_0^{T} dt}=\frac{F_0}{2}.$$

 

[sysprog]

I can now see how to get the 1/2 factor in. The impulse equation is $J=m\Delta v=\bar F \Delta t$. One has to cook up a force function $F(t)$ to describe a contact force parametrized by $F_0$ such that $$\bar F\equiv \frac{\int_0^{T} F(t)~dt}{\int_0^{T} dt}=\frac{F_0}{2}.$$

I think that this is insightful and helpful, but it seems to me that it doesn't directly address the question $-$ the title asks for help in formulating a word problem to which the equation specified (albeit probably partly incorrectly, due to the omission of kg that you postulated) is the correct answer; it doesn't ask for a more elucidative equation $-$ maybe a bat hitting a pitched ball would work, or if we want to set one side stationary, a golf club hitting a golf ball, or a bowling ball hitting a pin $-$ I also think that we haven't yet been given the complete exact original problem statement.

 

[kuruman]

I think that this is insightful and helpful, but it seems to me that it doesn't directly address the question $-$ the title asks for help in formulating a word problem to which the equation specified (albeit probably partly incorrectly, due to the omission of kg that you postulated) is the correct answer; it doesn't ask for a more elucidative equation $-$ maybe a bat hitting a pitched ball would work, or if we want to set one side stationary, a golf club hitting a golf ball, or a bowling ball hitting a pin $-$ I also think that we haven't yet been given the complete exact original problem statement.

It is true that we don't have the complete exact original statement. As it stands, one can think of a variety of physical situations that could fit the answer. Maybe the complete statement will narrow all these to one choice.