The Impulse

Misticles

Ive been working on the following equation for a while now and just cannot get it.

Its for wrking out the impulse experienced by a particle over time t.

∫〖Fx dt〗

=∫〖m dvx/dt〗 dt

= m(Vx/t – Vx/0)

= -2mVx

the x's are subscripts and the upper limit is t, lower limit 0.

I dont know where the -2mVx comes from at all!

Its been wrecking my brain for hours now!!

Please help me

Professor_Fink

Misticles wrote: »

Ive been working on the following equation for a while now and just cannot get it.

That's because you don't have enough information in what you have said to actually reach that answer:

Misticles wrote: »

∫〖Fx dt〗

=∫〖m dvx/dt〗 dt

= ∫ m dv

Misticles wrote: »

= m(V(t) – V(0)) (edited to fix notation)

So we're ok up to this point. However, you now need to know what the final velocity of the ball is, or some other related initial condition, which you have not mentioned. I can only assume that this equation was related to something bouncing off the floor/a wall. Without such context we simply don't know enough to progress beyond the above equation, since it depends on the change in velocities. However, if we are dealing with a collision with a fixed object, then the exiting velocity will be equal in magnitude to the initial velocity (to reach the equation you have you need them to be in opposite directions, as with a front on collision, as opposed to an oblique collision). This gives v(t) = -v(0)

This then trivially leads to

Misticles wrote: »

= -2mV

Misticles

Thanks.

Its relating to a particle bouncing elastically off a solid wall.

I still dont know where the 2 comes from?!

Theres not a mention of 2 anywhere in the notes.

Professor_Fink

Misticles wrote: »

I still dont know where the 2 comes from?!

Theres not a mention of 2 anywhere in the notes.

It comes from the last line. You have Impulse = mv(t) + mv(0). Let u = v(0), then since it leaves at the same speed in the opposite direction v(t) = -u. Thus you have Impulse = m(-u) - mu = -2mu.