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Bullet colliding with block problem (conservation of momentum & energy)

  1. Mar 10, 2010 #1
    1. The problem statement, all variables and given/known data
    An 8.00 g bullet is fired into a 250 g block that is initially at rest at the edge of a table of height 1 m. The bullet remains in the block, and after the impact the block lands 2 m from the bottom of the table. Determine the initial speed of the bullet.

    2. Relevant equations
    Conservation of Momentum (for perfectly inelastic conditions)
    m1v1(initial) + m2v2 (initial) = (m1+m2)vf
    m1 = mass of bullet (.008 kg)
    v1 = velocity of bullet
    m2 = mass of block (.250 kg)
    v2 = velocity of block (0 m/s)
    vf = final velocity of block and bullet right after collision

    Conservation of Energy
    KE + PE (after collision) = KE + PE (on ground), rewritten as:
    1/2mvi^2 + mgyi = 1/2mvf^2 + mgyf
    m = mass of bullet + block (.258 kg)
    vi = initial velocity of bullet + block after collision
    g = 9.8 m/s^2
    yi = 1 m
    vf = 0 m/s
    yf = 0 m

    3. The attempt at a solution
    I think I have this problem solved correctly, but wondered if someone could confirm that for me (those darn even-numbered problems at the back of the book without answers!).

    First, I use the conservation of momentum and put in my known values, ending up with:
    .008v1 = .258vf

    Next, I use the conservation of energy and put in the known values, ending up with 4.427 m/s for the initial velocity of the block and bullet, which si the final velocity in my first equation.

    I substitute 4.427 m/s into my first equation and get an answer for the initial speed of the bullet as 142.75 m/s.
  2. jcsd
  3. Mar 11, 2010 #2


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    vf = 0 m/s - this is wrong.

    The block+bullet performs projectile motion with horizontal initial velocity. What will be its velocity just before the impact with the ground?

  4. Mar 11, 2010 #3
    I thought that the final velocity for the conservation of energy equation would be 0 m/s because the block had come to rest? There was another example problem in my book with the same situation except the block was on a pendulum and swung upwards to a certain height, and they set the vf there to 0 m/s so I followed that example.

    I'm unclear on why the vf shouldn't be 0 m/s.

    However, I did solve to find the vf just before it impacts the ground and I got 6.13 m/s, which led me to a final answer of the speed of the bullet of 118.36 m/s.
  5. Mar 11, 2010 #4


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    The box will come to rest after the impact with the ground. It is an other story. Up to this impact, it accelerates vertically and its initial potential energy transforms to kinetic energy, while it retains the initial horizontal velocity. You need the horizontal velocity. Conservation of energy yields the vertical velocity, you can use it to calculate the time of flight. The horizontal distance where the box lands is 2 m. How do you get the horizontal velocity?

  6. Mar 11, 2010 #5
    I get the horizontal velocity by first finding the time i takes for the block to fall 1 m, in (delta)y = 1/2ayt^2 and get .452 seconds. Next I find the velocity in the x direction by using (delta)x = v(initial)t and get 4.24 m/s. I then use Vy(final) = vy(initial) + ayt and get vy = -4.43 m/s. I take the square root of 4.24 + -4.43 and get the total final velocity as 6.13m/s.

    If the box comes to rest after impacting the ground at 2 m, then I don't think they give me enough information to figure that out, because I would have to know exactly at what distance it came to rest or some kind of friction force, right? It seems to me they only give me enough information to work with the block stopping at 2 m.
  7. Mar 11, 2010 #6


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    That is almost correct. vx= 2/0.452=4.42 m/s.

    You calculated the final speed, but you do not need it. You do not need conservation of energy now.

    The horizontal velocity is constant during the fall. The block+bullet moves horizontally on the table just after the collision. So the velocity of the block+bullet is 4.24 m/s in the x direction after collision. .
    The momentum is a vector quantity, it holds separately for all components. The bullet moves horizontally (it is not stated, but you can assume, otherwise they should have given the angle), the block+bullet moves horizontally on the table till it falls down, so you can write the equation for conservation of the x component of the momentum :

    m(bullet)*vi(bullet)= (m(bullet)+m(block)) *vx and get the initial velocity of the bullet.

    The problem did not say that the block comes to rest at 2 m from the table. It was said that

    "after the impact the block lands 2 m from the bottom of the table." It is completely irrelevant what happens to the block after touching the ground.

  8. Mar 11, 2010 #7
    This is is the same answer I get for the initial velocity of the block when I use the KE + PE (after collision) = KE + PE (on ground) equation from my first attempt at a solution.

    This equation ends up being the same as the the equation I end up with after solving this equation from my initial post: m1v1(initial) + m2v2 (initial) = (m1+m2)vf

    Either equation gives me:
    .008vi = .258vx

    Where I can substitute 4.42 m/s for vx, getting:
    .008vi = .258(4.42)
    vi = 142.5

    This answer for the initial velocity of the bullet is the same as the solution in my first post.

    I guess I'm unsure as to why my first attempt at a solution is wrong. Did I just luck out in getting the same answer? I do understand what you mean about the vf = 0 m/s being wrong, though. I suppose the point is that I should solve this equation the correct way, not the way I did because it assumes vf = 0, which is incorrect.
  9. Mar 12, 2010 #8


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    Yes, the solution is the same, but such things do happen. You make a few mistakes, and accidentally you get a result which is the same as the correct one.

    You used this equation to get vi:

    1/2mvi^2 + mgyi = 1/2mvf^2 + mgyf

    yi=1 m, yf =0 (ground) vf=0, so 1/2mvi^2 + mg*1=0,

    that means negative vi^2 which is nonsense. But the square root of 2g is 4.42 - by accident, it is the same as the real value of the initial velocity.

    Try to repeat the calculation with a table of 1.5 m high.

  10. Mar 12, 2010 #9
    Ah you are so right!! Thank you so much for setting me straight on this. :)
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