# Im all confused

1. Jan 19, 2004

### photon

In the equation F=ma, "a"is for acceleration right?
I mean, acceleration before an object hits something doesnt affect force, but the velocity does as it hits, right?
So why not F=mv?
Or maybe Im missing something here.
[?] [?]

2. Jan 19, 2004

### chroot

Staff Emeritus
The equation "F = ma" says that to cause an object of mass m kilograms to accelerate at a meters per second per second, you must apply a force of F newtons. It does not deal with collisions directly.

- Warren

3. Jan 19, 2004

### jimmy p

yeah, in A-level maths, N2L (Newtons 2nd Law) which deals with the equation F=ma is ALWAYS used in exams to determine what the acceleration of a train with 2 carriages is, with a fixed mass, a driving force and obviously resistance. It is also used to work out the tension in towbars, i think it can be used to work with pulleys as well...

EDIT, the train is at rest.

Last edited: Jan 19, 2004
4. Jan 19, 2004

### Kalimaa23

What got you confused is the role the linear momentum plays in collisions.

Actually, the correct form for Newton's second law is : dp/dt = F, where p = mv is the momentum.

So, dp/dt = v.(dm/dt) + m(dv/dt) = m(dv/dt) = ma , if the mass is constant (if dm/dt = 0).

When a bal hits a wall, it's speed, and therefore it's momentum change. The change in momentum is then the force.

5. Jan 19, 2004

### LURCH

The force when an object hits something actually is acceleration. Like in Dimitri's example (a ball hitting a wall), the ball's velocity (relative to the wall) goes from whatever speed it was travelling to zero in a certain amount of time, because it can't happen instantaniously. Now, from the wall's frame of reference, this is a deceleration, but from the ball's frame of reference, it is acceleration. This is because once the ball is in motion and no longer accelerating, it can be seen as being "at rest" relative to its own inertial frame. Then this big old wall comes along and hits it, causing it to accelerate to match the wall's velocity.

So the ball's mass accelerates from zero to the wall's velocity in a very short period of time, generating a lot of force.

6. Jan 19, 2004

### Integral

Staff Emeritus
Another basic consideration. The units of Force are mass* Length* Time-2

The units of mass * Velocity is

Mass * Length * Time -1

Clearly mv cannot be a force, it does not have the correct units.

7. Jan 19, 2004

### Arcon

Impulse is an important quantity which describes such a collision. The change in momentum is related to the change in speed as

dp = m*dv

Since dp/dt = F ==> dp = F*dt

The impulse is then defined as the quantity integral (F*dt)

http://scienceworld.wolfram.com/physics/Impulse.html

Yes!!! I so love it when people say that! :-)

I wish people would get the idea out of their head that force is defined as F = ma when in fact, as Dimitri points out, force is defined as F = dp/dt.

Thanks Dimitri

8. Jan 20, 2004

### Kalimaa23

It is the duty of every physics major to root out these misconceptions.

Don't get me started on E = mc²...

9. Jan 20, 2004

### jimmy p

Sorry guys, i only know the simple stuff (Stupid A-Level physics) so to me at the moment, N2L is F=ma

10. Jan 20, 2004

### Arcon

It appears to me that people associate Newton's second law with F = ma rather than F = dp/dt because this is how they were first introduced to it in, say, highschool or before they learned calculus.

If one doesn't know calculus then one can't understand what F = dp/dt is. So how would a highschool text explain F = dp/dt to a class of highschool students who don't know calculus? They'd use the equality "F = ma" which does not require any understanding of calculus. That equality holds in all Newtonian applications for which mass is constant.

But F = ma is wrong. If you have the Feynman lectures then read about this in there. Feynman always does a great job at explaining this stuff.

11. Jan 20, 2004

### Kalimaa23

Well, now you know that it's not. Go and show off to your friends already !

12. Jan 20, 2004

### jimmy p

I assume by calculus you mean differentiation hence the dp/dt. Point taken, it does make sense and i think i learned something!

13. Jan 20, 2004

### photon

Thanks guys. I think I had the wrong idea about the equation.