Bullet striking block, maximum height obtained?

[JFonseka]

Homework Statement

A 10-g bullet moving 1000 m/s strikes and passes through a 2.0-kg block initially at rest, as shown. The bullet emerges from the block with a speed of 400 m/s. To what maximum height will the block rise above its initial position?

Homework Equations

K = (mv^2)/2
W = F x d
F = ma

The Attempt at a Solution

I first calculated the initial kinetic energy of the bullet which is 5000 J, and the final as it exits the block, which is 800J, therefore am I right in assuming 4200 J has been given to the block?

I tried using E = mgh for the block, then I realized I did not know how quickly the block accelerated upwards, and to find that I would need F = ma, but I don't know the force either, do i need to find those two to finish this question off? Or is there another way to go about it.

 

[learningphysics]

Can you describe the picture, or post it? I'm guessing the block is on an incline, and the bullet is moving horizontally and strikes the block?

This isn't an elastic collision... so energy is converted to heat here... so you can't use conservation of kinetic energy during the collision.

 

[JFonseka]

No the block is not on an incline, and I think heat energy produced here is negligible.

The bullet moves upwards, and the block is sitting on a ledge, the ledge has a small gap in which the bullet goes through and strikes the block

 

[JFonseka]

Pictures doesn't fit, edited

 

[learningphysics]

No the block is not on an incline, and I think heat energy produced here is negligible.

The bullet moves upwards, and the block is sitting on a ledge, the ledge has a small gap in which the bullet goes through and strikes the block

Do they give the height of the block? And also, is the 1000m/s right before it hits the block?

You should use conservation of momentum during the collision to calculate the velocity of the block right after the collision.

Then you can use conservation of energy (kinetic energy and grav. potential energy) to see the height it goes up to.

 

[learningphysics]

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There is a pitiful drawing :P

That's actually very good.

 

[JFonseka]

Do they give the height of the block?

You should use conservation of momentum during the collision to calculate the velocity of the block right after the collision.

Then you can use conservation of energy (kinetic energy and grav. potential energy) to see the height it goes up to.

No they don't give the height of the block, but thanks, I'll try momentum, forgot about that!

Cheers

 

[JFonseka]

That's actually very good.

Lol the drawing doesn't hold once it's posted, damn formatting

 

[JFonseka]

Yep, so I did mv + mv = mv + mv

Which means, 2(0) + 0.010(1000) = Momentum before collision = 10 kg/ms^-1

2(v) + 0.010(400) = Momentum after collision, leading to v = 3 m/s

K.E. of the block therefore is .5x2x9 = 9J

h = 9/19.62 = 0.46m, which is one of the answers, so I'm guessing it's right, thanks!

 

[learningphysics]

Yep, so I did mv + mv = mv + mv

Which means, 2(0) + 0.010(1000) = Momentum before collision = 10 kg/ms^-1

2(v) + 0.010(400) = Momentum after collision, leading to v = 3 m/s

K.E. of the block therefore is .5x2x9 = 9J

h = 9/19.62 = 0.46m, which is one of the answers, so I'm guessing it's right, thanks!

cool. you're welcome.