Find the speed of the bullet as it emerges from the block

[innightmare]

Homework Statement

A 10.0g bullet is shot through a 1.0 kg wood block suspended on a string 2.0m long. The center of mass of the block rises a distance of 1.0cm. find the speed of the bullet as it emerges from the block if its initial speed is 450m/s.

Homework Equations

I think this has to do with linear momentum, p(i)=p(f)(mv=mv). Either that or conservation of energy: mgh +1/2mv^2=mgy +1/2mv^2

The Attempt at a Solution

Im confused as to how to go about solving this problem. for the initial mv, I plugged in the values for the bullets mass and used 450 m/s as the initial velocity. for the final mv, i plugged in the the mass value for the block and bullet, and used that to divide my initial mv, which was good for linear momentum but still not good enough, i don't think its right. so when i use the conservation of energy,i know my h=2.0m and y=.01m

Am I on the right track. Which equation should i use? Thanks

 

[learningphysics]

The idea is to use:

mbullet*vi = mbullet*vf + mblock*vblock

where vf is the velocity of the bullet right after it goes through, and vblock is the velocity of the block right after it goes through.

so you need to find vf. but before you can do that you need vblock... do you see a way you can find the velocity of the block right after the bullet goes through?

 

[ogb p]

I would first clarify the assumptions and limitations of the problem. Is the bullet assumed to be a point object? Is there any air resistance? Is the string assumed to be massless? These details can affect the accuracy of the final answer.

Next, I would approach the problem by using the principle of conservation of momentum. This principle states that the total momentum of a system remains constant, unless external forces act upon it. In this case, the external force acting on the system is the bullet being shot through the block.

Using the equation p(i) = p(f), where p is the momentum, we can set up the following equation:

m(bullet)v(i) = (m(bullet) + m(block))v(f)

where m(bullet) is the mass of the bullet, v(i) is the initial velocity of the bullet, m(block) is the mass of the block, and v(f) is the final velocity of the bullet and block together.

Solving for v(f), we get:

v(f) = (m(bullet)v(i)) / (m(bullet) + m(block))

Plugging in the given values, we get:

v(f) = (0.01 kg * 450 m/s) / (0.01 kg + 1 kg) = 4.5 m/s

Therefore, the speed of the bullet as it emerges from the block is approximately 4.5 m/s.

As a final step, I would also check if this answer makes sense intuitively. Since the block is suspended on a string, it is likely that the bullet would not emerge from the block with a very high velocity. The answer we have calculated seems reasonable in this context.