1. In terms of impulse and momentum, why are air bags in automobiles are a good idea?
dp/dt = F; where dp is the change in momentum over interval dt. For a given change in momentum (e.g., during a crash), longer dt as stretched out by the air bags means less force F acting on the body. dp is fixed at dp = m(v1 - v0) where v1 is the initial car speed before the crash and v0 = 0 after the crash. m is the body mass protected by the air bag.
Impulse = dp = Fdt; so the air bags increase the dt and decrese the F for a given dp.
2. You can't throw a raw egg against a wall without breaking it, but you can throw it at the same speed ito a sagging sheet without breaking it. Why?
Same as 1. The dt to slow down the egg from v1 to v0 is lengthened; so the force is reduced. m is the egg mass.
3. A moving car has a mometum. If the car moves twice as fast, what is its momentum. Now consider two cars, one twice as heavy as the other, both of which are moving down a hill at the same speed. Compare to the lighter car, what is the momentum of the heavier car?
p = mv; P = mV = m2v where m is the car mass and V = 2v is twice as fast. P/p = m2v/mv = 2; so that P = 2p going twice as fast doubles the momentum. p = mv and P = Mv = 2mv; where M = 2m the car is twice as massive. Thus P/p = 2mv/mv = 2; so that P = 2p once again.
4. What does it mean to say that momentum is conserved?
It is neither created nor destroyed. Example, two perfectly inelastic balls m and M are going at each other at V and v velocity respectively. Before colliding the momenta of the two balls are p = mV + Mv; after collision, they're stuck on each other and the momentum is now P = (m + M)v', what is the resulting velocity v' of the two stuck balls?
Because momentum is conserved, we can write p = mV + mv = (m + M)v' = P as momenta before the collision is the same as momentum after it (nothing added; nothing subtracted...it's conservered). Now v' = (mV + Mv)/(m + M) to find the resulting velocity of the two stuck balls.
5. Image that youa re hovering next to the space shulttle in an Earth's orbit. YOur buddy of equal mass, who is moving at 4km/h with respect to the shuttle, bumps into you. If he holds onto you, how fast do you both move with respect to the ship?
Set M = m and go back to 4. v' = m(v + V)/2m = (v + V)/2 As I'm still before the collision, you must set v = 0 and V = 4 kph. You can do the math.
6. Jocko, who has a mass of 60kg and stand at rest on ice, catches a 20kg ball that is thrown to him at 10km/h. How fast does Jocko and the ball move across the ice?
Again, go back to 4. Set M = 60 kg, m = 20 kg, V = 10 kph, and v = 0 before the catch (collision). Solve for v'.