Bullet fired into a block that and swings like a pendulum

[kerbyjonsonjr]

Homework Statement

A 2.5 kg wood block hangs from the bottom of a 1.0 kg, 1.2 m long rod. The block and rod form a pendulum that swings on a frictionless pivot at the top end of the rod. A 12 g bullet is fired into the block, where it sticks, causing the pendulum to swing out to a 33 angle.

Homework Equations

KE=1/2mv²
gravitational potential=mgh

The Attempt at a Solution

Since the kinetic energy of the bullet all ends up as potential energy at the top of the swing I set kinetic energy equal to the sum of all the gravitational potential energies. The height that the rod goes would be from its center of mass so it would be cos33=h/.6 which equals .5m. Then the height of the bullet and the block would be cos33=h/1.2 which equals 1 m. Then I used the equation 1/2mv₂= mgh + (m+m)gh and that plugged in with numbers looks like 1/2(.012)v²=1(9.81)(.5) +(2.5 +.012)(9.81)(1) and then I solved for v and got v= 70.2 m/s

I do not know where I went wrong. If anybody could help me that would be great and I would really appreciate it.

 

[tiny-tim]

hi kerbyjonsonjr!

this is a perfectly inelastic collision … energy is not conserved, and you have to use conservation of momentum or angular momentum (momentum and angular momentum are always conserved in collisions) …

only after you find the initial speed can you then use conservation of energy

 

[Doc Al]

Since the kinetic energy of the bullet all ends up as potential energy at the top of the swing

Ah, but it doesn't. The collision between bullet and block is perfectly inelastic. Mechanical energy is not conserved during the collision, but something else is. What?

You'll need to treat the behavior of bullet and pendulum in two phases:
(1) The collision itself.
(2) The rising of the pendulum after the collision.

Different things are conserved during each phase.

 

[44r0n0wnz]

Al is right. In the first part, energy is not conserved because the collision is inelastic. After the collision, the block moves right after the bullet stops, making this part of it elastic. Solve by using conservation of momentum for the first part and conservation of energy for the second part.

 

[rwisz]

I would like to point out that your solution is not entirely accurate. While your equations and calculations are correct, there are a few assumptions and simplifications that may affect the accuracy of your results.

Firstly, the assumption of a frictionless pivot may not be entirely realistic. In real-world scenarios, there will always be some level of friction present which can affect the motion of the pendulum.

Secondly, the pendulum is not a perfect system. The block and rod will experience some level of deformation upon impact from the bullet, which can affect the conservation of energy and the overall motion of the pendulum.

Thirdly, air resistance is not taken into account in your solution. In real-world scenarios, the bullet will experience air resistance as it travels through the air, which can affect its speed and kinetic energy upon impact with the block.

To improve the accuracy of your solution, you can consider incorporating these factors into your equations and calculations. This may involve using more complex equations and taking into account variables such as friction, deformation, and air resistance. Additionally, performing experiments to measure and verify the results can also help to improve the accuracy of your solution.

Overall, while your approach is correct, it is important to keep in mind the limitations and assumptions in your solution and to continuously strive for more accurate and realistic results in scientific research.