# A bullet ricochets off a steel plate problem

• Deckname
In summary: To get the duration of the change (Δt), assume the frictional force is constant and find the time for the given drop in tangential velocity.
Deckname
Homework Statement
What's the median resisitive force the plate offers against the bullet given the following info? (in the picture attached)
Relevant Equations
none have been given, most likely ##F_{avg}=\frac{\Delta p}{\Delta t}##
I have calculated KE_i and KE_f, took the difference between the initial and final kinetic energy of the bullet to be equal to the work spent to overcome the friction, and divided it by the distance traveled, but arrive at around 20000N. The solution should be 9.5*10^7.

Not sure what else to try at this point. Suggestions or hints please? Also this is a part of my second year mechanics seminar, and I wouldn't be asking for the solution or anything here unless passing the subject literally depended on it, as missing even 1 solution out of the 29 exercises I've already solved, it's still counted a fail. So please lend a hand!

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your handwritten stuff is (1) not really acceptable for this forum and (2) not readable even if it were.

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Alright, thanks for pointing that out, I won't post handwritten stuff again. But then again I've explained what I've done in the OP (forgot to add I divided the work by distance, since F=W/d) and I've also attached a picture of the problem itself that I made in paint to be clearly readable.

What is the change in the bullet’s momentum?

Nugatory said:
What is the change in the bullet’s momentum?
Alright, I now realize I'm supposed to calculate the solution through F_avg=ΔP/Δt, but how do you calculate the change in momentum when not only the momentum changed, but also it's direction? I'm confused now.

I've used the ##F_{avg}=\frac{\Delta P}{\Delta t}## and have arrived at the same result of ~##20000N##. I have calculated ##\Delta t## by using ##v^2={v_0}^2+2a \Delta x## to calculate the acceleration and plugged it in ##t=\frac{\Delta v}{a}## to get time. I calculated ##\Delta p## by taking into consideration the velocity vectors parallel to the trajectory of the bullet, since this is the change in momentum we want to consider, and I then plugged all of that into the first equation I mentioned, ##F_{avg}##, and got ##20000N## again.

I'm actually confused now, am I constantly making a mistake in all the different ways this can be solved, since I keep arriving at the almost identical result (different but a few hundreds of N), or is the result my professor has provided wrong?

Deckname said:
I've used the ##F_{avg}=\frac{\Delta P}{\Delta t}## and have arrived at the same result of ~##20000N##. I have calculated ##\Delta t## by using ##v^2={v_0}^2+2a \Delta x## to calculate the acceleration and plugged it in ##t=\frac{\Delta v}{a}## to get time. I calculated ##\Delta p## by taking into consideration the velocity vectors parallel to the trajectory of the bullet, since this is the change in momentum we want to consider, and I then plugged all of that into the first equation I mentioned, ##F_{avg}##, and got ##20000N## again.

I'm actually confused now, am I constantly making a mistake in all the different ways this can be solved, since I keep arriving at the almost identical result (different but a few hundreds of N), or is the result my professor has provided wrong?
You are still misunderstanding change in momentum. Momentum is a vector, obtained by multiplying mass by the velocity vector. The change in momentum is the change in the vector. You are only looking at change in speed and energy, which are scalars.
E.g. in an elastic bounce there is no change in speed or energy but maybe a substantial change in velocity and momentum.

To get the duration of the change (Δt), assume the frictional force is constant and find the time for the given drop in tangential velocity.

## 1. How does a bullet ricochet off a steel plate?

When a bullet hits a steel plate, it transfers its kinetic energy to the plate. The steel is able to withstand the force of the bullet and deform slightly, but ultimately the bullet bounces off at an angle due to the conservation of momentum.

## 2. What factors affect the angle of ricochet?

The angle of ricochet is affected by the angle at which the bullet hits the steel plate, as well as the angle and shape of the plate itself. The type of bullet and the speed at which it is traveling can also impact the angle of ricochet.

## 3. Can a bullet ricochet multiple times off a steel plate?

Yes, it is possible for a bullet to ricochet off a steel plate multiple times if the angle of impact and the angle of the plate are just right. However, each ricochet will decrease in speed and energy, making subsequent bounces less likely.

## 4. Can a bullet ricochet off other materials besides steel?

Yes, a bullet can ricochet off a variety of hard surfaces such as concrete, stone, or even water. However, the likelihood and angle of ricochet may vary depending on the material's hardness and the angle of impact.

## 5. Is there a way to predict the angle of ricochet?

There are mathematical formulas and computer simulations that can estimate the angle of ricochet based on the properties of the bullet, the material it is hitting, and the angle of impact. However, there are many variables involved, so the exact angle of ricochet may be difficult to predict accurately.