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Momemtum/conservation of momentum

  1. Aug 20, 2005 #1
    Can anyone explain to me in simple terms about momentum and conservation of momentum. I understand the terms, but don't understand how it relates to a smaller object sliding sideways into a non-moving larger object. Before the smaller object hits the larger, doesnt it have a larger momentum? How can the smaller object transfer its momentum to a larger object which has mass but no velocity? I am assuming that something small hitting something bigger would not conserve any momentum because of their differences. :confused:
  2. jcsd
  3. Aug 20, 2005 #2
    momentum is always conserved. consider if a fly hat weighs .01 lbs with vel of 5 mph hits a truck that weighs 5 tons that is at rest. what would be the outcome?
  4. Aug 20, 2005 #3


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    " I am assuming that something small hitting something bigger would not conserve any momentum because of their differences."

    Since that completely contradicts what any textbook would tell you why would you assume it?

    Yes, a "small object" moving at any speed has greater momentum (in magnitude at least) than a larger object motionless because the large object has none at all. However, remember that momentum is a vector quantity. If you are working in one dimension- motion on a single line- you still have to remember that momentum may be postive or negative. So:

    Suppose an object of mass 1kg is moving at 2 m/s. It collides with a motionless object of mass 2 kg.
    Before the collision, the total momentum is (1 kg)(2 m/s)+ (2kg)(0 m/s)= 2 kg m/s. Let v1 be the velocity of the 1 kg mass and v2 be the velocity of the 2 kg mass after the collision. Then because momentum is conserved, we must have (1 kg)(v1)+ (2kg)(v2)= 2 kg m/s.

    Notice that that is one equation in 2 unknowns. One way for it to be satisfied is to have v1= 0 (the first mass stops completely) and v2= 1 m/s.

    Yet another would be for the two masses to "stick" together having the same velocity: v1= v2= v so (1 kg)(v)+ (2kg)(v)= (3kg) v= 2 kg m/s so v= 2/3 m/s.

    The reason there are an infinite number of solutions is that we haven't considered the kinetic energy ((1/2)mv2) involved. If this were a "perfectly elastic" collision, kinetic energy would also be conserved so we would have (1/2)(1kg)v12+ (1/2)(2kg)(v22)= (1/2)(1 kg)(4 m2/s2 or
    v1+ v2= 4 kg m2/s2= 4 Joules.

    Now solve the two equations v1+ 2v2= 2 and
    v12+ v22= 4 simultaneously.

    From the first equation, v1= 2- 2v2. Putting that into the second equation, 4- 8v2+ 4v22+ v22= 4 or v22- 822= 0. That has two solutions: v2= 0 and v2= 8 m/s.

    The first of those would be where the moving mass misses and there really isn't any collision at all! Putting v2= 8 into the first equation we have v1+ 2(8)= 2 or v1= -16 m/s.

    That's how a small mass makes a larger one move: it bounce back the other way at high speed!
  5. Aug 20, 2005 #4
    Lets take a special case of the smaller mass hitting the larger one and then stopping.

    In this case it has transferred all its momentum to the larger mass.

    The larger mass will now move. But with a speed slower than the smaller ball initially had.

    Momentum => M_(initial) * Velocity_(initial) = M_(finall) * Velocity_(final).

    Or a situation could arise where the smaller ball continues with a lesser speed in another direction and the larger ball also begins to move but with an even smaller speed.

    That is... If you sum up the momentum before a collision and then afterwards they will be equal.

    Momentum is always conserved!

    Though this is not the case for kinetic energy.

  6. Aug 20, 2005 #5
    Momentum is a vector quantity defined as the mass of an object multiplied by its velocity. Momentum is conserved when there is no net external force acting on the system. Conservation simply means constant in time.

    look at the momentum for the two objects? One has momentum the other does not. Which is which? Why does one have momentum while the other doesn't?


    Through the collision. When they touch, momentum is transferred. This is a good question.

    Well it all depends if there are any external forces present. Are there? It depends on the set-up. In a typical textbook set-up we usually assume that there are no external forces unless we are explicitly told so.
    Let us say that we have a comet of mass m1 moving at a velocity v1i towards a stationary asteroid of mass m2. We are in a universe where they are the only 2 objects. Let us also assume both are spheres and the collision is head on. What is the velocity of asteroid after the collision if we measure the velocity of the comet to be v1f in the same direction as v1i after the collision?

    use conservation of momentum.
    [tex] \vec{p}_{1i} + \vec{p}_{2i} = \vec{p}_{1f} + \vec{p}_{2f} [/tex]

    Which means:

    [tex] m_1 \vec{v}_{1i} + m_2 \vec{v}_{2i} = m_1 \vec{v}_{1f} + m_f \vec{v}_{2f} [/tex]


    [tex] v_{2i} = 0 [/tex]

    Which means that the initial momentum of the asteroid is zero also.

    now just solve for the final velocity of the asteroid:

    [tex] \vec{v}_{2f} = \frac{m_1}{m_2} ( \vec{v}_{1i} - \vec{v}_{1f} )[/tex]
  7. Aug 20, 2005 #6
    This is untrue.
    Momentum is only conserved if there is no net external force acting on the system.
    \Sigma_i \vec{F}_i = \frac{d\vec{p}}{dt}
    \Sigma_i \vec{F}_i = 0
    \frac{d\vec{p}}{dt} = 0
    which means
    [tex] \vec{p} = constant [/tex]
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