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Math Help - Physics Problems: Conservation of Momentum

  1. #1
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    Question Physics Problems: Conservation of Momentum

    I'm completely stumped on a few problems that are due after this weekend. If anyone could offer me hints on how to start them, it would be great. Thanks!

    1. A 50 g bullet lodges into a 2.0 kg block of clay hung by a string. The bullet and clay rise together to a height of 12 cm. What was the velocity of the 50 g mass just before entering?

    2. A 0.150 kg bullet is fired at 715 m/s into a 2.0 kg wooden block at rest. The velocity of the block afterward is 40 m/s. The bullet passes through the block and emerges with what velocity?

    3. A 2.0 kg ball moving to the right at 1.0 m/s strikes a 4.0 kg ball moving left at 3.0 m/s. What are the velocities after impact, assuming complete elasticity?

    4. A 2.0 kg block A and a 1.0 kg block B are pushed together against a spring and tied with a cord. When the cord breaks, the 1.0 kg block moves to the right at 8.0 m/s. What is the velocity of the 2.0 kg block?



    Edit: It would appear the OP now only requires help with Questions 1. and 3.
    Last edited by mr fantastic; February 21st 2009 at 01:37 PM. Reason: Restored the original questions that were deleted by the OP in an edit
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  2. #2
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    Quote Originally Posted by xxlvh View Post
    I'm completely stumped on a few problems that are due after this weekend. If anyone could offer me hints on how to start them, it would be great. Thanks!

    1. A 50 g bullet lodges into a 2.0 kg block of clay hung by a string. The bullet and clay rise together to a height of 12 cm. What was the velocity of the 50 g mass just before entering?

    2. A 0.150 kg bullet is fired at 715 m/s into a 2.0 kg wooden block at rest. The velocity of the block afterward is 40 m/s. The bullet passes through the block and emerges with what velocity?

    3. A 2.0 kg ball moving to the right at 1.0 m/s strikes a 4.0 kg ball moving left at 3.0 m/s. What are the velocities after impact, assuming complete elasticity?

    4. A 2.0 kg block A and a 1.0 kg block B are pushed together against a spring and tied with a cord. When the cord breaks, the 1.0 kg block moves to the right at 8.0 m/s. What is the velocity of the 2.0 kg block?
    Some random thoughts to be applied (after due thought) where appropriate:

    Think about conservation of mechanical energy after impact and think about conservation of momentum during impact.

    Elastic collision means kinetic energy of the system is conserved.
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    So for the elastic collision the total sum of the momentum and kinetic energy must be equal after the equation. I don't know how to set up the equation to solve though, since none of the final velocities were given..
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    Quote Originally Posted by xxlvh View Post
    [snip]3. A 2.0 kg ball moving to the right at 1.0 m/s strikes a 4.0 kg ball moving left at 3.0 m/s. What are the velocities after impact, assuming complete elasticity? [snip]
    Quote Originally Posted by xxlvh View Post
    So for the elastic collision the total sum of the momentum and kinetic energy must be equal after the equation. I don't know how to set up the equation to solve though, since none of the final velocities were given..
    I don't see the trouble here. The final velocities are the things you're trying to find. So you give them a symbol like u and v and substitute into the appropriate equations.

    2(1) + 4(-3) = 2u + 4v .... (A)

    \frac{1}{2} (2) (1)^2 + \frac{1}{2} (4) (3)^2 = \frac{1}{2} (2) u^2 + \frac{1}{2} (4) v^2 .... (B)

    I'll let you figure which equation corresponds to which conserved quantity, and you should obviously simplify the equations. Then solve equations (A) and (B) simultaneously.
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    Thank you, I'll work on finishing that question up then... any suggestions for problem #1?
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    1. A 50 g bullet lodges into a 2.0 kg block of clay hung by a string. The bullet and clay rise together to a height of 12 cm. What was the velocity of the 50 g mass just before entering?

    Energy should be conserved - I hope. So:
    Kinetic E of bullet = Potential E of {bullet+block}
    1/2 * m_b * V^2 = (m_b+m_Block)*g*height

    Find V above.

    Be careful about the conversion of units... g, kg, cm, m, etc...
    and unit of g=m/s^2
    Last edited by mr fantastic; February 22nd 2009 at 01:47 AM. Reason: Merged posts and edited
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