Conservation of Energy Problem -- bullet fired into a ballistic pendulum

In summary, the approximate velocity of the bullet fired into a ballistic pendulum, with a mass of 7.5 g, is 25.6 m/s in an inelastic collision. In order to conserve momentum, the velocity of the bullet and block together is found to be 1.4 m/s. Using the conservation of momentum, the velocity of the bullet alone is then calculated to be 470 m/s. The conservation of kinetic energy is not applicable in this scenario due to the inelastic collision.
  • #1
logan3
83
2

Homework Statement


A bullet has a mass of 7.5 g. It is fired into a ballistic pendulum. The pendulum's receiving block of wood is 2.5 kg. After the collision, the pendulum swings to a height of 0.1 m. What is the approximate velocity of the bullet?

[itex]m_{bullet} = 7.5g = 0.0075 kg[/itex]
[itex]m_{wood} = 2.5 kg[/itex]
[itex]h_i = 0 m[/itex]
[itex]h_f = 0.10 m[/itex]
[itex]g = 9.8 m/s^2[/itex]

Homework Equations


[itex]KE_i + PE_i = KE_f + PE_f[/itex]
[itex]KE_i = \frac {1}{2}m_{bullet}v_{bullet}^2[/itex]
[itex]PE_i = 0[/itex]
[itex]KE_f = 0[/itex]
[itex]PE_f = (m_{bullet} + m_{block})gh_f[/itex]

The Attempt at a Solution


[itex]KE_i + PE_i = KE_f + PE_f \Rightarrow \frac {1}{2}m_{bullet}v_{bullet}^2 + 0 = 0 + (m_{bullet} + m_{block})gh_f[/itex]
[itex]\Rightarrow \frac {1}{2}m_{bullet}v_{bullet}^2 = (m_{bullet} + m_{block})gh_f[/itex]
[itex]\Rightarrow v_{bullet} = \sqrt {\frac {2(m_{bullet} + m_{block})gh_f}{m_{bullet}}} = \sqrt {\frac {2((0.0075 kg) + (2.5 kg))(9.8 m/s^2)(0.10 m)}{(0.0075 kg)}}[/itex]
[itex]= 25.5986 \frac {m}{s} \sim 26 \frac {m}{s}[/itex]
 
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  • #2
Kinetic energy is not conserved when the bullet hits the block. There is another quantity that is conserved.
 
  • #3
mfb said:
Kinetic energy is not conserved when the bullet hits the block.
Why not?

Thank-you.
 
  • #4
@logan3 Kinetic energy is not conserved because this is not an elastic collision. This site explains it well, I think:

http://hyperphysics.phy-astr.gsu.edu/hbase/elacol.html

When two objects in a collision stick together, you can be sure it's an inelastic collision, not an elastic collision (in which the relative velocity between the objects is the same before and after). However, elastic or inelastic, there's another quantity always conserved in collisions.
 
  • #5
Thank-you, that's helpful. I'll think about it a bit longer.
 
  • #6
Edit: totally wrong post I was supposed to answer something else. Ignore.
 
  • #7

The Attempt at a Solution


Ok, the conservation of kinetic energy is not conserved between elastic and inelastic collisions, but momentum is. So first I will find the velocity in the inelastic collision, then use the conversation of momentum to solve for the velocity of the bullet.

Since I am studying the kinetic energy of the objects after an inelastic collision, then the equations should be:
[itex]KE_i = \frac {1}{2}(m_{bullet} + m_{block})v_{bullet+block}^2[/itex]
[itex]PE_i = 0[/itex]
[itex]KE_f = 0[/itex]
[itex]PE_f = (m_{bullet} + m_{block})gh_f[/itex]

[itex]KE_i + PE_i = KE_f + PE_f \Rightarrow \frac {1}{2}(m_{bullet} + m_{block})v_{bullet+block}^2 = (m_{bullet} + m_{block})gh_f[/itex]
[itex]\Rightarrow \frac {1}{2}v_{bullet+block}^2 = gh_f \Rightarrow v_{bullet+block} = \sqrt{2gh_f} = \sqrt{2(9.8 m/s^2)(0.10 m)} = 1.4 \frac {m}{s}[/itex]

Now, since momentum is conserved between elastic and inelastic collisions, I can use that to relate the two velocities to each other ([itex]v_{bullet}[/itex] and [itex]v_{bullet+block}[/itex]).

[itex]\rho_{bullet} + \rho_{block} = (m_{bullet} + m_{block})v_{bullet+block}[/itex]
[itex]\Rightarrow m_{bullet}v_{bullet} + m_{block}v_{block} = (m_{bullet} + m_{block})v_{bullet+block}[/itex]
[itex]\Rightarrow v_{bullet} = \frac {(m_{bullet} + m_{block})v_{bullet+block} - m_{block}v_{block}}{m_{bullet}} = \frac {(2.5075 kg)(1.4 m/s) - 0}{0.0075 kg} = 468.06 \frac {m}{s} \sim 470 \frac {m}{s}[/itex]
 
  • #8
That looks correct.
 

Related to Conservation of Energy Problem -- bullet fired into a ballistic pendulum

What is conservation of energy and how does it relate to a bullet fired into a ballistic pendulum?

Conservation of energy is a fundamental law in physics that states energy cannot be created or destroyed, but can only be transformed from one form to another. In the case of a bullet fired into a ballistic pendulum, this law states that the total energy of the system (bullet + pendulum) must remain constant before and after the collision.

How does the ballistic pendulum work?

The ballistic pendulum is a device used to measure the speed of a projectile, such as a bullet. When the bullet is fired into the pendulum, it transfers some of its kinetic energy to the pendulum, causing it to swing. By measuring the height of the pendulum swing, the initial speed of the bullet can be calculated.

What factors affect the conservation of energy in this problem?

Several factors can affect the conservation of energy in this problem, including the mass and velocity of the bullet, the mass and height of the pendulum, and any external forces acting on the system. Friction and air resistance can also play a role in reducing the total energy of the system.

Why is conservation of energy important in this problem?

Conservation of energy is important in this problem because it allows us to accurately determine the speed of the bullet by measuring the height of the pendulum swing. Without this law, the total energy of the system could change, making it difficult to calculate the initial speed of the bullet.

What are some real-world applications of this problem?

The principles of conservation of energy and the ballistic pendulum are used in various fields, including forensic science, ballistics, and engineering. It can be used to determine the speed of a bullet in a crime scene investigation, to test the performance of firearms, and to study the effects of collisions in different materials.

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