# Basic momentum and magnitude questions!

1. Dec 9, 2009

### Ronaldo21

1. A student wearing frictionless in-line skates on a horizontal surface is pushed by a friend
with a constant force of 47 N. How far must the student be pushed, start-
ing from rest, so that her final kinetic energy is 354 J ?

for this i did 354/2 which i got 177 then i divided by 47 . is that correct?

2 (part 1 of 2) A 4.78 g bullet moving at 510.8 m/s penetrates a tree trunk to a depth of 4.93 cm. a) Use work and energy considerations to find the magnitude of the force that stops the bullet.

for this i did KE=1/2mv^2 and i found the answer and did f=ma. is that right??

3. (part 2 of 2) b) Assuming that the frictional force is constant, determine how much time elapses between the moment the bullet enters the tree and the moment the bullets stops moving.

for this one i just did vf^2=vi^2+2at. is that correct??

THANK YOU!!

2. Dec 9, 2009

### Cryxic

The first part is wrong. You need to use the following formula:

$$Work = Force * Distance$$

So it should just be 354/47, giving you 7.5 meters.

I can't tell from what you've written if you're right on part 2 or not, so here's how you do it. You need to set the kinetic energy of the bullet equal to the work done to stop it, which you can relate to force -- the answer you're looking for. Mathematically:

$$\frac{1}2mv^2 = W = Fd$$

You know the mass and the velocity of the bullet. You know the distance d it travels inside the trunk. Solve for F and you should get about 12,650 newtons (but check my work).

For part 3....knowing the force it took to stop the bullet, you can figure out the acceleration from F = ma. Once you know the acceleration, you put it into the following kinematics equation:

$$x = v_it + \frac{1}2at^2$$

where t is the time you need to solve for. You know x (distance bullet travels inside tree), you know initial velocity (510.8 m/s), and you know the acceleration that you calculated above. Solve for time t, but you will need the following quadratic equation:

$$1323106.69t^2 + 510.8t - .0493 = 0$$

You will get two values: 8 x 10^-5 seconds or -4.66 x 10^-4 seconds. You have to dismiss the second value on physical grounds (can't have negative time). So you should get get about 8 x 10^-5 seconds. Again, check my work, but that answer makes sense intuitively. The bullet should stop really quickly given the enormous force opposing its motion.

3. Dec 9, 2009

### Ronaldo21

ohh i see. okay i understand part 1 completly but not part 2.
so i have to find time right? where did you get 1323106.69 from like i got 2646213. hmm. im confused.lol

4. Dec 9, 2009

### Cryxic

It's just divided by two to get a simpler constant in front of the t^2.