Calculating Bullet Speed from Impact with Lumber

[shann0nsHERE]

A 35.0-g bullet strikes a 4.7-kg stationary piece of lumber and embeds itself in the wood. The piece of lumber and bullet fly off together at 8.0 m/s. What was the original speed of the bullet?
I don't know where to begin...

 

[shann0nsHERE]

This one too pleeasee

A thread holds a 1.5-kg and a 4.50-kg cart together. After the thread is burned, a compressed spring pushes the carts apart, giving the 1.5 kg cart a speed of 27 cm/s to the left. What is the velocity of the 4.5-kg cart?



I would be so grateful if someone could help me..

 

[rl.bhat]

Just go through the conservation of linear momentum.

 

[Saladsamurai]

Hi there. Is conservation of linear momentum something you have studied in class?

Also: I would start a new thread for each problem, else things get confusing/messy.

 

[shann0nsHERE]

Yes, I am in the ib program but i was absent for that lesson..
ha thanks for the heads up, i just started this
Andd thank you both for the help

 

[Saladsamurai]

Yes, I am in the ib program but i was absent for that lesson..
ha thanks for the heads up, i just started this
Andd thank you both for the help

Okay, well paraphrasing Conservation of momentum says that if no external forces act on a system (which is true in this case), then the total linear momentum is conserved.

That is, \sum mv_{initial}=\sum mv_{final}

In this case,

M_{bullet}V_{bullet}+M_{block}V_{block}=M_{(block+bullet)}V'_{(block+bullet)}

where I used the ' symbol to indicate velocity after the collision

And Yes, they are both conservation of momentum.