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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?

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2 Answers 2

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.

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but we can't find velocity without time, so I guess we need to know time anyway (if we could find time and velocity, we can find acceleration also.) –  w4j3d Mar 13 '12 at 0:48
    
Say that the velocity is constant before impact, and we read the car's speedometer. –  Qmechanic Mar 13 '12 at 0:59
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@w4j3d we can find velocity without time if we use impulse. Impulse gets rid of the icky business of time and allows you to use conservation of momentum without interference from the time factor (which we don't know at times) –  Manishearth Mar 13 '12 at 1:12
    
@w4j3d: Also, here acceleration may not be constant, so we can't apply the SUVAT equations (v=u+at and all that). –  Manishearth Mar 13 '12 at 1:13
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Whenever the force is not easy to calculate. Impulse is $$J=\Delta p= \int\limits_0^{\delta t}F\mathrm dt$$. This comes directly from $F=\frac{\mathrm dp}{\mathrm dt}$

For most collisions, the force isn't easily defined. It's usually an extremely large force $F$ operating in an extremely short time $\delta t$. It need not be, though--most spring-mass collisions have a well-defined force. But for collisions like clapping your hands together, throwing a wooden block at a wall, etc, we can't easily calculate the force since we don't know a whole host of things, like the moduli of the object and material. So what we do is this: We safely neglect $\delta t$. During the colission, we have little or no change in displacement. The only thing that changes is velocity, so we can just code it as a change in momentum $J=\Delta p$. This way, we can work with the equations of conservation of momentum etc without having to calculate the collision force.

Some examples

Lets say you have a pulley system. A simple one: A person A is hanging from one end, and a block B of equal mass is on the other end. The system is at rest. An object C approaches A at a speed $v$. A catches it. Find the final velocities of the system.

Here, we cannot directly conserve momentum as the pulley hangs from a ceiling and that ceiling can take away momentum. Neither can we conserve energy as the collision is inelastic. But, using impulses, we can conserve momentum: Let $J_T=\int Tdt$ where $T$ is the tension. Since tension is same in both strings attached to A and B, this impulse will act on both. It is also upwards for both, as tension is upwards. Now, we can actually conserve momentum individually for the two bodies. Instead of using $\Delta p=0$, we use $J= \Delta p$. Your $J$ can be eliminated easily (two equations, two unknowns--$J$ and final velocity)

Another example is when you have a ball bouncing off a wall; when the wall has friction. We can apply normal coefficient of restitution equations to find the final perpendicular velocity $v_\perp$, but we can't do the same for the parallel direction, as friction is there. But, we know that $f=\mu N$. Using this, we get $J_f=\mu J_N$ (plug it into $\int Fdt$, $\mu$ is a constant). Now, since we already calculated the change in velocity, we can calculate $J_N=J_\perp=m\Delta v_\perp$. Using this, we can find $J_f$. Since $J_f=J_\parallel=\Delta v_\parallel$, we can find the change in parallel velocity. That's where we can get these formulae from.

If you want elaboration on the examples, please let me know. I've assumed a bit while using them.

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"Lets say you have a pulley system. A simple one: A person A is hanging from one end," I enjoyed reading the answer, until I read this peace of horror, specially that it's complemented with "and a block B of equal mass is on the other end.". Good, the person is dead in equilibrium. –  w4j3d Mar 13 '12 at 5:00
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@w4j3d =P By 'hanging', I didn't mean he was executed (I'm quite sure that dead men catch no objects, even if in equilibrium).. I meant he was hanging-from-a-rope-like-tarzan sort of thing. If you want to avoid that, just make him sit in a large scale pan... Hypothetical situations are always strange. –  Manishearth Mar 13 '12 at 5:09
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protected by Qmechanic Jan 10 at 23:36

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