One dimensional inelastic collision; bullet through block.

In summary, the problem involves a 12.8 g bullet moving upward at 930 m/s, passing through the center of mass of an 8.3 kg block at rest, and emerging with a velocity of 520 m/s. The maximum height of the block above its initial position can be found using the equation m1v1i + m2v2i = m1v1f + m2v2f, with the bullet's final velocity becoming the block's initial velocity. The mistake in solving for the maximum height was due to incorrect algebra, but it was resolved by realizing that the acceleration due to gravity should be negative.
  • #1
J-dizzal
394
6

Homework Statement


In the figure here, a 12.8 g bullet moving directly upward at 930 m/s strikes and passes through the center of mass of a 8.3 kg block initially at rest. The bullet emerges from the block moving directly upward at 520 m/s. To what maximum height does the block then rise above its initial position?

20150707_211326_zpsglwxmosy.jpg


Homework Equations


m1v1i + m2v2i = m1v1f + m2v2f

The Attempt at a Solution


20150707_210809_zpsywi6ntir.jpg

not sure where i went wrong with this one.
 
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  • #2
J-dizzal said:

Homework Statement


In the figure here, a 12.8 g bullet moving directly upward at 930 m/s strikes and passes through the center of mass of a 8.3 kg block initially at rest. The bullet emerges from the block moving directly upward at 520 m/s. To what maximum height does the block then rise above its initial position?

[ IMG]http://i1164.photobucket.com/albums/q562/falsovero/20150707_211326_zpsglwxmosy.jpg[/PLAIN]

Homework Equations


m1v1i + m2v2i = m1v1f + m2v2f

The Attempt at a Solution


[ IMG]http://i1164.photobucket.com/albums/q562/falsovero/20150707_210809_zpsywi6ntir.jpg[/PLAIN]
not sure where i went wrong with this one.
Velocity for block is correct.

Bad algebra in solving for y-y0 .
 
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  • #3
Instead of y-y0 I will call it h and instead of a, I will call it g.

You wrote the equation v2=2gh... good so far... but then you wrote h=v2-2g :confused:
 
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  • #4
Nathanael said:
Instead of y-y0 I will call it h and instead of a, I will call it g.

You wrote the equation v2=2gh... good so far... but then you wrote h=v2-2g :confused:
I need to take an algebra class, again.
 
  • #5
why is g positive in this problem? because the bullet is fired straight up and so the accerleration of g is -9.8m/s/s. I can see it works out in the equation for it to be positive, i just don't know the reason why.
 
Last edited:
  • #6
Thank you
 
  • #7
J-dizzal said:
is g positive here? or would it be negative because it in the negative y direction?
So I'm assuming that you resolved the above question, because of your following post.
J-dizzal said:
Thank you
It may be helpful to you and us, if you could learn to use the edit feature and/or the delete feature to alter or to remove a post like Post #5.
 
  • #8
SammyS said:
So I'm assuming that you resolved the above question, because of your following post.

It may be helpful to you and us, if you could learn to use the edit feature and/or the delete feature to alter or to remove a post like Post #5.
Sure, i didnt know about the delete feature. I'll edit it above.
 
  • #9
J-dizzal said:
why is g positive in this problem? because the bullet is fired straight up and so the accerleration of g is -9.8m/s/s. I can see it works out in the equation for it to be positive, i just don't know the reason why.
Because it is really Vf (which you called V) that is zero, and the V0 is what is 0.6323. So g would be negative but you would also have to subtract V02 over to the other side and then the negative sign would cancel out.
 
  • #10
Nathanael said:
Because it is really Vf (which you called V) that is zero, and the V0 is what is 0.6323. So g would be negative but you would also have to subtract V02 over to the other side and then the negative sign would cancel out.
Ok, i think i get it now because the problem is treated as two problems and the v_f i got from the first problem becomes v0 in the next. Thanks.
 

What is a one dimensional inelastic collision?

A one dimensional inelastic collision is a type of collision where two objects collide in a straight line and stick together after the collision. In this type of collision, kinetic energy is not conserved and is lost due to the deformation of the objects involved.

What happens during a one dimensional inelastic collision?

During a one dimensional inelastic collision, the two objects involved stick together and move with a common velocity after the collision. This is because the kinetic energy is not conserved and is converted into other forms of energy, such as thermal or sound energy.

What is the difference between a one dimensional inelastic collision and a one dimensional elastic collision?

In a one dimensional elastic collision, both momentum and kinetic energy are conserved. This means that the two objects involved bounce off each other after the collision with no loss of kinetic energy. In a one dimensional inelastic collision, only momentum is conserved and kinetic energy is lost.

How is the coefficient of restitution related to a one dimensional inelastic collision?

The coefficient of restitution is a measure of the elasticity of a collision. In a one dimensional inelastic collision, the coefficient of restitution is equal to 0, indicating a completely inelastic collision where the two objects stick together. In contrast, a one dimensional elastic collision would have a coefficient of restitution equal to 1, indicating a perfectly elastic collision where there is no loss of kinetic energy.

How can we calculate the final velocity of the two objects after a one dimensional inelastic collision?

The final velocity of the two objects after a one dimensional inelastic collision can be calculated using the conservation of momentum principle, which states that the total momentum before the collision is equal to the total momentum after the collision. This can be represented by the equation m1v1 + m2v2 = (m1 + m2)vf, where m1 and m2 are the masses of the two objects, v1 and v2 are their initial velocities, and vf is their final velocity after the collision.

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