Piston/bullet inelastic collision (not like other problems)

In summary, a bullet with a mass of 0.01kg and a velocity of 150m/s strikes and embeds itself into a piston in a cylinder filled with 5.0m^3 of air at STP. The piston has a mass of 0.1kg and a surface area of 10cm^3. The system is considered adiabatic after the collision. The goal is to find the displacement of the piston and the time period of oscillation. To do so, we can use the equations for momentum and kinetic energy before and after the collision, as well as the formula for time period of oscillation. We can also consider the adiabatic process and the relationship between pressure and
  • #1
Kyle2016
3
0

Homework Statement


a bullet has mass=0.01kg velocity=150m/s, you have a cylinder with 5.0m^3 of air at STP with a piston at one end that has mass=0.1kg and surface area of 10cm^3 and is at rest. The bullet strikes and embeds itself into the piston.
The system is to be considered adiabatic after the collision

a)find the displacement of the piston

b)what is the time period of oscillation of the piston

Homework Equations


momentum, p=mv momentum of an inelastic collision= m1v1+m2v2= (m1+m2)V(final)
kinetic energy of the system before collision=(1/2)[mass(bullet)][velocity(bullet)]^2
kinetic energy of the system after collision=(1/2)[mass(bullet)+mass(piston)][velocity(final)]^2
F=ma, F=kx
time period of linear oscillation= 2(pi)sqrt([mass(bullet)+mass(piston)]/k)

The Attempt at a Solution


I calculated the final velocity- V(final)=[0.1*150]/[0.1+0.01]= 13.63636363...m/s

I calculated the kinetic energy before- (1/2)[0.1][150]^2=112.5 joules
I calculated the kinetic energy after-(1/2)[0.1+0.01][13.636363...]^2=10.2272 joules
Kinetic Energy lost=102.2728 joules <-this energy lost is imparted to the gas in the cylinder as heat(that is what we are told)
I know that a=(kx)/m
omega-w=sqrt(k/m) there for a=sqrt(k/m)[x/m]

I know that the gas in the cylinder exerts a force on the piston and this force will slow the pistons movement till it stops and then the force will over come it and push the piston out till a "vacuum" force pulls the piston back in and thus the linear oscillation begins. I can't for the life of me figure out how to find "x" or "k" for that matter.

The only other hint we were given is that air is to be considered diatomic for this problem and thus Cv=5/2 R and Cp=7/2 R thus gamma can be calculated as 7/5
but I don't know how this plays into it. Any insight would be helpful I have been staring at this and googling for at least 8 hours now.
 
Physics news on Phys.org
  • #2
Kyle2016 said:

Homework Statement


a bullet has mass=0.01kg velocity=150m/s, you have a cylinder with 5.0m^3 of air at STP with a piston at one end that has mass=0.1kg and surface area of 10cm^3 and is at rest. The bullet strikes and embeds itself into the piston.
The system is to be considered adiabatic after the collision

a)find the displacement of the piston

b)what is the time period of oscillation of the piston

Homework Equations


momentum, p=mv momentum of an inelastic collision= m1v1+m2v2= (m1+m2)V(final)
kinetic energy of the system before collision=(1/2)[mass(bullet)][velocity(bullet)]^2
kinetic energy of the system after collision=(1/2)[mass(bullet)+mass(piston)][velocity(final)]^2
F=ma, F=kx
time period of linear oscillation= 2(pi)sqrt([mass(bullet)+mass(piston)]/k)

The Attempt at a Solution


I calculated the final velocity- V(final)=[0.1*150]/[0.1+0.01]= 13.63636363...m/s

I calculated the kinetic energy before- (1/2)[0.1][150]^2=112.5 joules
I calculated the kinetic energy after-(1/2)[0.1+0.01][13.636363...]^2=10.2272 joules
Kinetic Energy lost=102.2728 joules <-this energy lost is imparted to the gas in the cylinder as heat(that is what we are told)
I know that a=(kx)/m
omega-w=sqrt(k/m) there for a=sqrt(k/m)[x/m]

I know that the gas in the cylinder exerts a force on the piston and this force will slow the pistons movement till it stops and then the force will over come it and push the piston out till a "vacuum" force pulls the piston back in and thus the linear oscillation begins. I can't for the life of me figure out how to find "x" or "k" for that matter.

The only other hint we were given is that air is to be considered diatomic for this problem and thus Cv=5/2 R and Cp=7/2 R thus gamma can be calculated as 7/5
but I don't know how this plays into it. Any insight would be helpful I have been staring at this and googling for at least 8 hours now.

You have a piston and a cylinder filled with air. The bullet strikes the piston and drives it into the cylinder. What happens to the air inside the cylinder?

The clue is in this statement, "The system is to be considered adiabatic after the collision." What's the relationship between pressure and volume in a cylinder which can be considered adiabatic, i.e., there is no heat transfer to or from the cylinder to its surroundings?

https://en.wikipedia.org/wiki/Adiabatic_process
 
  • Like
Likes Kyle2016
  • #3
Consider P when the piston is displaced by some distance x . At that instant , what is the pressure ?

You will get P as a function of x , and thus F as a function of x .

So use a kinematic equation → a = v*(dv/dx) and solve from there .

Hope this helps .
 
  • #4
SteamKing said:
You have a piston and a cylinder filled with air. The bullet strikes the piston and drives it into the cylinder. What happens to the air inside the cylinder?

The clue is in this statement, "The system is to be considered adiabatic after the collision." What's the relationship between pressure and volume in a cylinder which can be considered adiabatic, i.e., there is no heat transfer to or from the cylinder to its surroundings?

https://en.wikipedia.org/wiki/Adiabatic_process

Okay this helped alot... I can see that
6e0bc8d5bc392b6a685e362884024348.png

This would allow me to find P2 and thus V2, V2-V1 would tell me the change to give me x

The part that I'm stuck on is how to find T2. I know that I have 102.2728 joules of energy creating the temperature change thus I would be tempted to use Q(heat energy in joules)=mcΔT, but air's specific heat capacity changes with temperature and pressure thus how can I calculate delta T without knowing the P2 at which the specific heat capacity would be applied. Perhaps, I should just assume that the specific heat capacity is 1kJ/kg*K since the changes to the specific heat are rather small. Sorry for the questions. I think I might be over thinking this...
 
  • #5
Do not use relation between P and T ; use the relation between P and V .
 
  • #6
Qwertywerty said:
Do not use relation between P and T ; use the relation between P and V .

The only other equation I know is ((P1V1)^gamma)/T1=((P2V2)^gamma)/T2) this was provided by our professor because he said you have to account for the temp change but this still leaves me with too many unknowns.
 
  • #7
The basic equation for an adiabetic process is P*(V)∧γ = constant .
 
  • #8
Kyle2016 said:
The only other equation I know is ((P1V1)^gamma)/T1=((P2V2)^gamma)/T2) this was provided by our professor because he said you have to account for the temp change but this still leaves me with too many unknowns.

Are you sure about this equation ?
 

What is a piston/bullet inelastic collision?

A piston/bullet inelastic collision is a type of collision between a piston and a bullet where the kinetic energy of the system is not conserved and is lost as heat and sound.

How does a piston/bullet inelastic collision differ from other collision problems?

In other collision problems, such as a perfectly elastic collision, the kinetic energy of the system is conserved. However, in a piston/bullet inelastic collision, the kinetic energy is not conserved and is instead lost as heat and sound.

What factors affect the outcome of a piston/bullet inelastic collision?

The outcome of a piston/bullet inelastic collision can be affected by factors such as the mass and velocity of the piston and bullet, the material properties of both objects, and the angle at which they collide.

Why is it important to study piston/bullet inelastic collisions?

Piston/bullet inelastic collisions are commonly seen in firearms and understanding the physics behind these collisions is important for improving firearm design and safety. Additionally, this type of collision can also be seen in other industries such as automotive and manufacturing.

How can the energy lost in a piston/bullet inelastic collision be minimized?

The energy lost in a piston/bullet inelastic collision can be minimized by using materials with high elasticity, reducing the angle of collision, and controlling the velocity and mass of the objects involved. Additionally, implementing proper lubrication and maintenance can also help reduce energy loss in these collisions.

Similar threads

  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
749
  • Introductory Physics Homework Help
Replies
7
Views
3K
  • Introductory Physics Homework Help
Replies
28
Views
2K
Replies
1
Views
583
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
1K
Back
Top