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

  • Thread starter logan3
  • Start date
67
2
1. The problem statement, all variables and given/known data
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]

2. Relevant 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]

3. 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]
 
32,577
8,463
Kinetic energy is not conserved when the bullet hits the block. There is another quantity that is conserved.
 
67
2
Kinetic energy is not conserved when the bullet hits the block.
Why not?

Thank-you.
 
@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.
 
67
2
Thank-you, that's helpful. I'll think about it a bit longer.
 
Edit: totally wrong post I was supposed to answer something else. Ignore.
 
67
2
3. 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]
 
32,577
8,463
That looks correct.
 

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