What about the more common glancing blow Remember

What about the (more common) glancing blow? Remember: contact forces are NORMAL forces! To understand what effect this has, consider bouncing a ball off the ground…

A glancing blow with the ground is like partly traveling straight down into it and, at the same time, trying to travel past it.

Normal Force Weight The unbalanced (net) force is parallel to the inclined surface of the road! Exactly the direction we know a car left in neutral will start to coast!

A glancing blow with the ground is like partly traveling straight down into it and, at the same time, trying to travel past it. The horizontal This part of its motion (the downward component of its motion is of its velocity) is reversed unaffected! by the collision, just like a head-on collision event.

Remember: contact forces are NORMAL forces! The component heading right toward the center of mass is reversed by the blow! While the perpendicular component continues unchanged! Here’s the direction it rebounds!

mv Before After Two billiard balls undergo the glancing (not head-on) collision shown above (some spin must have been placed on the cue ball, but you don’t have to worry about that detail). What is the direction of the target ball after the collision? This, one of the final pieces of momentum, plus what other piece will still total to the same momentum we started with?

When an object explodes or breaks apart: : Is this ever possible (one half remains at rest behind…the other flies off in one direction)?

Deep in free space? Why?

Explosion Before the explosion: Mass, M Initially: vo = 0 P=0 because v 0=0 After the explosion: v 1 v 2 m 1 Finally: P = m 1 v 1 + m 2 v 2 which still must = 0 which means: m 1 v 1 = -m 2 v 2

A stationary uranium nucleus undergoes radioactive fission. Before fission: Uranium nucleus After fission: 1 2 Which fragment has a greater momentum? A) 1 B) 2 C) both the same speed?

p=0 pgas procket pi = 0 = pf = pgas + procket pgas = – procket pi = 0 = pf = prifle + pbullet prifle = – pbullet

A cannon rests on a railroad flatcar with a total mass of 1000 kg. When a 10 kg cannon ball is fired at a speed of 50 m/sec, as shown, what is the speed of the flatcar? A) B) C) D) 0 ½ 1 20 m/s to the right m/s to the left m/s to the right

1000 kg flatbed car at rest, loaded with 40 kg of ammo and carrying a 90 kg rifleman. The rifleman fires a 4 gram bullet at 965 m/sec straight out the back of the car. The entire car recoils how fast? initial momentum 0 final momentum mbulletvbullet + mcarvcar = -0. 0034 m/sec = -3. 4 mm/sec But what about by the time he has used up his last shell, firing his 10, 000 th time?

A bullet’s muzzle velocity (the speed it leaves the rifle with and races away from its rifle barrel) is 400 m/sec But did I mention the muzzle is onboard a flatbed car traveling 20 m/sec? 20 m/sec

50 m/sec A cannon rides a railroad flatcar, traveling to the right at 50 m/sec. It then fires a 10 kg cannon ball with a muzzle velocity of 50 m/sec to the left. What horizontal speed do observers stationary on the ground observe for this bullet? A) 0 m/sec (it simply drops straight down to the ground). B) 25 m/sec. C) 50 m/sec. D) 75 m/sec. E) 100 m/sec.

50 m/sec A cannon rides a railroad flatcar, traveling to the right at 50 m/sec. It then fires a 10 kg cannon ball with a muzzle velocity of 50 m/sec to the left. What affect does this have on the speed of the flatcar? A) The car speeds up. B) The speed of the car does not change. C) The car slows down.

50 m/sec mballvball + Mcarvcar = mballv'ball + Mcarv'car the ball leaves the muzzle both had the same with v=0 in the initial velocity = 50 m/sec earth’s reference frame (mball+ Mcar)50 m/sec = Mcarv'car ( 1010 kg )50 m/sec = (1000 kg)v'car = 50500 kg·m/sec / 1000 kg v'car = 50. 500 m/sec

? A bomb at rest explodes into four fragments. The momentum vectors for three of the fragments are shown. Which arrow below best represents the momentum vector of the fourth fragment?

? No external forces act on the bomb, so its momentum must be conserved: the total momentum before the explosion is zero, so total momentum after must also be zero.

An explosive charge separates rocket stages high in earth’s atmosphere. Which best represent the trajectories of the stages? A B C

An artillery shell bursts at the peak of its trajectory. Which best represents its streaming fragments? A C B D

SOME ANSWERS Question 1 will still total to the same momentum we started with? This, one of the final pieces of momentum, plus what other piece Which fragment has a greater momentum? Question 2 C) both the same Question 3 speed? A) 1 Same momentem! Total momentum before fission is zero, so total after must still be zero (no external forces so momentum is conserved). Since momentum is a vector, fragments 1 and 2 must have equal and opposite momenta. Since fragment 1 has a smaller mass, it must have greater speed since v = p / m. Question 4 B) ½ m/s to the right Question 5 A) 0 m/sec (it simply drops straight down to the ground). Question 6 A) The car speeds up. As shown on the slide that follows this question. No external forces act on the bomb, so its momentum must be conserved: the total momentum before the explosion is zero, so total momentum after must also be zero. Question 7 Question 8 C Question 9 D The explosive charge both slows down the discarded fuel stage (dropping its trajectory) and speeds up the capsule (raising its trajectory). You may be used to watching fireworks, which look an awful lot like A. However those are shot straight up and timed to got off near the peak of their trajectory.