Bullet conservation of momentum problem

[BrainMan]

Homework Statement

A 7-g bullet is fired into a 1.5-kg ballistic pendulum. The bullet emerges from the block with a speed of 200 m/s, and the block rises to a maximum height of 12 cm. Find the initial speed of the bullet.

Homework Equations

conservation of momentum

The Attempt at a Solution

First I tried to find the initial momentum of the bullet as if it did't penetrate the block
so .007v_i = (1.507)v_f
because I have two unknowns I used the conservation of mechanical energy to solve for v_f
1/2(1.507)(v_f²) = (1.507)(9.8)(.12)
v_f = 1.54 m/s
then plug that into find the initial velocity
v_i = 435.29 m/s
initial momentum = 3.05

Then I tried to find the momentum of the bullet exiting the block and add it to the initial momentum to get the total initial momentum.
200(.007) = 1.4
1.4 + 3.05 = 4.45
and finally I divided that by the mass to get the initial velocity 635.75 m/s. The correct answer is 529 m/s.

 

[Nathanael]

v_i = 435.29 m/s

How did you get this number?

0.007v_i = (1.507)v_f
...
v_f = 1.54 m/s

From this? (If yes, then double check your numbers.)

 

[Nathanael]

Other than that, looks good!

P.S.
You should not round your numbers until the very end. (Try to solve the problem with just algebra and then plug in the numbers at the end.) The reason is that your answer will be slightly off.

(In this problem and another you've recently posted, you rounded before the end which caused your answers to be off by about half a percent, which isn't much, but in some problems rounding before the end will cause you to be off by more than half a percent, and in other problems half a percent could be a meaningful error.)

 

[BrainMan]

Other than that, looks good!

P.S.
You should not round your numbers until the very end. (Try to solve the problem with just algebra and then plug in the numbers at the end.) The reason is that your answer will be slightly off.

(In this problem and another you've recently posted, you rounded before the end which caused your answers to be off by about half a percent, which isn't much, but in some problems rounding before the end will cause you to be off by more than half a percent, and in other problems half a percent could be a meaningful error.)

OK I see what I did! Thanks!

 

[HalfLostAndSearching]

Your approach of using conservation of momentum and energy is correct. However, there are a few errors in your calculation.

1. In your first attempt, you used the wrong mass for the block. The mass of the block is given as 1.5 kg, not 1.507 kg.

2. In your second attempt, you added the momentum of the bullet after exiting the block to the initial momentum. This is incorrect because the bullet is no longer in contact with the block and therefore does not contribute to the momentum of the system.

3. The final calculation for the initial velocity in your second attempt is incorrect. The correct calculation is (4.45 m/s) / (0.007 kg) = 635.71 m/s, not 635.75 m/s.

To get the correct answer of 529 m/s, you can use the following steps:

1. Calculate the initial momentum of the bullet using the given information: (0.007 kg)(vi) = (1.5 kg)(0 m/s) + (0.007 kg)(200 m/s)

2. Use conservation of energy to find the final velocity of the block: (1/2)(1.5 kg)(vf^2) = (1.5 kg)(9.8 m/s^2)(0.12 m)

3. Use conservation of momentum to find the final velocity of the bullet: (0.007 kg)(vi) = (1.5 kg)(vf)

4. Substitute the final velocity of the bullet into the equation from step 1 to solve for vi.

5. The final answer should be vi = 529 m/s.