forcesmomentumnewtonian-mechanics
In my notes it says
"The ideas of impulse and momentum is useful in solving problems where:-
a) the force F is not easily calculable (e.g. sudden impact or blow)
b) the impulse force is so large that the body suffers deformation (i.e. the impact is no longer elastic)
There is no explanation for what is said. When I think about it, force is the rate of momentum. How/why could we calculate momentum for small amount of time (sudden impact), but not calculate force?
Unless I made some mistakes, impulse is equal to momentum and not to change of momentum. Where did I go wrong, or, what is the final word?
I'm not sure I completely understand your notation in your reasoning preceding the statements quoted above but 3rd line certainly doesn't look correct:
Acceleration = a=[1]Δv change of velocity (m) v/s
The correct equation is
$$\bar a = \frac{\Delta v}{\Delta t} $$
where $\bar a$ is the average acceleration. With this correction, the final equation becomes
$$J = (m)\bar a \Delta t = (m) \Delta v = \Delta p$$
This seems so straightforward that I suspect I don't understand what you're actually trying to show.
what is change of velocity then? if a football is at rest and I kick it and it aquires v=10m/s, haven't I accelerated it over a period of time? isn't that difference of Δv=+10m/s acceleration?
Yes, you accelerated it over a period of time and no, the difference in velocity is not acceleration.
(Average) acceleration is, as I wrote above, the ratio of the change in velocity to the elapsed time over which the change occurred.
So, if the change in velocity is
$$\Delta v = 10 \frac{m}{s}$$
one does not know the average acceleration unless one also knows the elapsed time $\Delta t$ since the average acceleration is given by
$$\bar a = \frac{\Delta v}{\Delta t}$$
Clearly, the average acceleration is inversely proportional to the elapsed time so, the smaller the elapsed time, the larger the average acceleration.
To be concrete, let us say that the foot was in contact with the football for $100 ms$. Then, the average acceleration of the football is
$$\bar a = \frac{10 \frac{m}{s}}{0.1s} = 100 \frac{m}{s^2} $$
The pulley (and the attachment to the ceiling) are part of the system here. Because of this, you cannot simply use conservation of momentum on the three given masses.
If the final velocity were $v$, then the total energy of the system would have increased since both the pan and counterweight would be moving and the other mass would not have slowed.
You can't use conservation of momentum equations on only a portion of a system. If you imagine a ball bouncing on the floor, you can't say the momentum of the ball is conserved before and after the bounce. You have to consider the change in momentum of the floor as well.
In your problem, the change in momentum of the ceiling will be small, but relevant.
$$\Delta p_{m1} + \Delta p_{m2} + \Delta p_{pan} + \Delta p_{ceiling} = 0$$
As you do not know the change in this final component, you can't use conservation to solve for the remaining momentum of the other three masses.
Let's change the situation to make this more explicit. Instead of a counterweight, consider two pans and two weights.
Let's imagine the pulley and string to be massless, so the two pans and two weights have a total mass of $4m$. If the balls have a velocity $v$, then the total momentum of the pulley inside the room is $1mv + 1mv = 2mv$.
But by symmetry, we can see that the pulley isn't going to turn. If we imagine the pans at rest after the collision, we find the momentum is now $0mv$.
If the pulley's connection to the ceiling/room/earth is not part of the system, then we say that forces from that connection were external and changed the total momentum. We cannot use conservation of momentum due to external forces.
If the ceiling/room/earth are part of the system, then after the collision, they have gained $2mv$ downward momentum, so the total system does not change. If we picture the box as nearly massless and in a spaceship instead of earth, the entire box would be moving downward at $v/2$ after the balls hit the pans. (assuming completely inelastic collision). The more massive the box, the slower it moves to retain the velocity. Consider it attached to a building/earth, and the momentum still changes, but the velocity change is no longer measurable.
Best Answer
Example: If a car with mass $m=1000 kg$ and with velocity $v=30m/s$ crashes into a wall (say without applying the breaks) and comes to rest, it is easy to calculate the total change in momentum
$$\Delta p~=~0-mv~=~\int F(t) dt,$$
also known as the impulse, but it is difficult to know the precise force $F(t)$ on the car as a function of time $t$ during the impact as the front of the car (and the wall) folds up in some particular way.