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Conservation of energy questions
By Peter Warner (P3272) on Sunday, November 12, 2000 - 12:54 am:

Hi,

I am stuck on the following Energy question, ie to use K.E. Could some one help me out, also if you can put a few comments regarding the reasoning of your method in terms of the energy. It would be most helpful. Thanks.

1. A bullet of mass 20g travelling at 500metres/sec horizontally hits a vertical wall. The wall exerts a constant resistance of 36000N on the bullet. Calculate the distance the bullet will penetrate the wall.

I also have a another question similar to this which is also troubling me, but if someone can help me out here, I'll try the other one myself.

Thanks.
Pete

By Susan Langley (Sml30) on Sunday, November 12, 2000 - 01:00 am:

OK, whats the initial KE of the bullet, use 0.5mv2. Make sure you convert the g into kg for SI units. OK do you know:
Energy=Force×Perpendicular displacement
So you put the KE of the bullet, which you work out, and the force into this equation to find the displacement

Hope that helps

Susan

By William Astle (Wja24) on Sunday, November 12, 2000 - 01:11 am:

I.
The rate of change of the total energy in an isolated system is zero. In other words energy is conserved. If energy can not flow out of or into a system then the total energy in the system is the same at any time.

II.
The kinetic energy of a body with mass m and speed u is (1/2)mv2. The work done on a body (energy imparted to it) by another body (for simple problems. In general it is better defined in terms of an infinite sum) is the product of the force applied to the body and the distance through which the force is applied.

So

1.
Initial Kinetic energy of the bullet is ½×20g×(500ms-1)2 = 2500J

The bullet does work on the wall and in doing so imparts energy of 2500J (since it eventually comes to rest and energy is conserved). So using the formula:

Work Done = Distance of Application × Force of Application

We have:

Distance of Application = Work Done/(Force of Application)

So here:

Distance = 2500J/36000N

which is approximately 7 cm.

By Hal 2001 (P3046) on Sunday, November 12, 2000 - 02:05 pm:

Thanks William, Susan.

Thanks for the method and also commentry on the conservation of evergy. Because of your help, I was able to tackle the other question (similar to this one, the other question gave me the amount the bullet went through the wall instead of giving me the force exerted and had to find that out). Thanks!!

Pete

By Peter Warner (P3272) on Monday, November 13, 2000 - 04:16 pm:

Hi,

I am trying to do this question using and considering Impulse formula and Energy (KE & PE) formula and not the constant acceleration formula. Can someone help out on part (b) of this question.

A ball of mass 0.4Kg is dropped from a height of 2.8m. After hitting the ground it rebounds to a height of 1.8m. By modelling the ball as a particle find:

(a) the speed with which the ball hits the ground
(b) the speed with which the ball rebounds from the ground
(c) the impulse the ball recieves from the ground


(a)
KE gained = 0.5mv2 = 0.4×0.5×v2

PE lost = mgh = 0.4×2.5×g

PE lost = KE gained ==> v = 7 metres/sec

(b)
i did this using the constant acceleration formula, but how can do this using energy?

u=? v=0 s=1.8 a=-g
using v2 = u2 + 2as ==> u = 5.93 metres/sec

(c)
m = 0.4, v = 5.93, u = 7
Impluse = mv -mu = 5.18 Ns


Thanks if you can help with doing part (b) using energy, or impulse/ momentum.

By William Astle (Wja24) on Monday, November 13, 2000 - 04:34 pm:

You know the potential of the ball when it reaches 1.8m. This is all the energy the ball has. Assuming it has not lost any in rising it will have the same energy (but all kinetic) just after it rebounds.

Use (1/2)mv2 to find the speed.

By Hal 2001 (P3046) on Monday, November 13, 2000 - 06:05 pm:

Hi William,
Thanks, it works!

Pete