Center of Mass and Linear Momentum?

Question

It is well known that bullets and other missiles fired at Superman simply bounce off his chest. Suppose a gangster sprays Superman's chest with 3 g bullets at the rate of 100 bullets/min, and the speed of each bullet is 500 m/s. Suppose too that the bullets rebound straight back with no change in speed. What is the magnitude of the average force on Superman's chest from the stream of bullets? Ans 5 N

Answer 1

Here's a hint:

One definition of "Force" is: "the rate of change of momentum."

This means: Average force is (change in momentum) divided by time.

So: By how much does each bullet's momentum change when it bounces off? That'll give you the "change in momentum" part.

And: How often does this happen? That'll give you the "time" part.

Answer 2

Change in momentum of each bullet
= 3e-3 kg * 2*500 m/s = 3 kgm/s

There are 100 of these per minute or 5/3 per second

Force = 3 kg m/s * 5/3 /s
= 5 N

Answer 3

In this kind of problems, air resistance is ignored. From your data, both fragments have the same vertical velocity (0), so they will reach ground at the same time (as well as the center of mass). But the center of mass moves as if no explosion took place (conservation of momentum) and reaches ground (assumed flat) at distance d. First fragment drops directly, from top of trajectory, so it travels d/2. Since the center of mass travels distance d, it follows that the second fragment travels distance 3d/2.

Answer 4

this is from Resnick and Halliday why don't you take a look at their online solution. search.