Bullet & 2 Blocks Momentum Question

[lando45]

Hey,

I have been set this question and thought I'd solved it correctly but it turns out I got the answers wrong...

A 5.50 g bullet is fired horizontally at two blocks at rest on a frictionless tabletop. The bullet passes through the first block, with mass 1.20 kg, and embeds itself in the second, with mass 1.80 kg. Speeds of 0.630 m/s and 1.40 m/s, respectively, are thereby given to the blocks. Neglect the mass removed from the first block by the bullet. Find the speed of the bullet immediately after it emerges from the first block, and find it's original speed.

For the first question (finding the speed after it emerges from the first block) I found 1/2mv2 for the second block then subtracted the 1/2mv2 of the first block and then calculated v and got a value of 20m/s but this is wrong...can anyone shed some light?

 

[Fermat]

Use conservation of momentum rather than conservation of energy.

 

[quicknote]

Hello,

It appears that you have made a mistake in your calculation. When calculating the speed of the bullet after it emerges from the first block, you should take into account the mass of the bullet (5.50 g) and the mass of the first block (1.20 kg). This will change the calculation from 1/2mv2 to (m1v1 + m2v2)/(m1 + m2), where m1 is the mass of the bullet and m2 is the mass of the first block. This should result in a speed of approximately 1.23 m/s, which is the correct answer.

As for the original speed of the bullet, you can use the conservation of momentum equation (m1v1 = (m1 + m2)v2) to find the initial speed of the bullet. This will give you a value of approximately 5.08 m/s, which is the correct answer.

I hope this helps clarify the problem for you. Remember to always double check your calculations and take into account all relevant masses in a system when solving momentum problems. If you continue to have trouble, feel free to ask for further clarification. Good luck!