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Homework Help: Challenging momentum questions

  1. Apr 27, 2004 #1
    Feedback any qs that are do able. Thnks.

    1. A child in a boat throws a 5.40kg package out horizontally with a speed of 10m/s. Calculate the velocity of the boat immediately after, assuming it was initially at rest. Say the mass of the child is 26kg and the mass of the boat is 55kg.

    2. A 1000kg Toyota collides with the rear end of a 2200kg Mercedes stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 2.8m before stopping. The police officer, knowing that the coefficient of kinetic friction between the tires and the road is 0.40, calculates the speed of the Toyota at impact. What was the speed?

    3. A softball of mass 0.220kg that is moving with a speed of 5.50m/s collides head on and elastically with another ball initially at rest. Afterwards it is found that the incoming ball has bounced backward with a speed of 3.7m/s. Calculate the mass of the target ball.

    4. A distant solar system has a central star (akin to our Sun) that is currently "burning out". It is projected that the star will decrease in mass by 5% over the next 20 years. Over the same time interval, the star will shrink in radius by 12%. By what percentage will the stars escape velocity increase in the next 20 years.
  2. jcsd
  3. Apr 27, 2004 #2
    1. Straight application of the law of conservation of momentum.

    2. Work backwards. Start with an energy argument and combine it with a conservation of momentum argument.

    3. Combine conservation of momentum with conservation of energy. Remember that an elastic collision is one in which kinetic energy is conserved.

    4. Use an energy argument to calculate the escape velocity. Then just replace m with .95m and r with .88r.

    Give them another shot and come back with what you've tried.

  4. Apr 27, 2004 #3
    The reader might assume from your response that energy is conserved in Question 2. Of course such is not the case. To be clearer you should probably mention that you use the work-energy theorem on the post-collision portion of the problem to find the speed of the combined system immediately after the collision. Then simply apply conservation of momentum to the collision itself to solve the rest of the problem.
  5. Apr 27, 2004 #4

    Cookiemonster's remarks are wonderfully clear, AS ALWAYS.
  6. Apr 27, 2004 #5
    Haha, thanks for the vote of confidence, holly. =]

  7. Apr 27, 2004 #6
    Okay, for #1, I would think that cons. of momentum of the whole crew = conservation of momentum for the thing thrown.

    P = mv, so for the thing thrown, P=5.4kg times 10 m/s, giving 54 kg m/s, I guess.

    Now, maybe cookie or someone could give advice here, but I guess the mass of the "crew & boat" would include the thing thrown out later, so it would be 26 kg plus 55 kg boat, plus the 5.4 kg item thrown.

    I am not sure we should be throwing the boat in, as it is floating, and Doc Al once took my head off for not being able to understand exactly what floating means concerning weight...

    But if we do add up the whole thing, it's 86.4 kg.

    so velocity for the boat without the item, once it was thrown...maybe v= momentum/mass? v=54 kg m/s divided by 86.4 kg?

    For the balls, maybe do a Case 1 and a Case 2...?
    Case one, P = mv...P=.220kg times 5.5 m/s...get that answer
    Case two, Above Answer = m times 3.7 m/s...divide each side by the 3.7 to get mass?

    I am not sure about that one.

    Good luck!!! Show some of your work/thinking, or the various Brains won't answer much for you...
  8. Apr 27, 2004 #7
    holly's got a neat idea, but unfortunately she's running in circles. =\

    The boat and child and package system was initially moving at a whopping 0 m/s. So its initial momentum is a whopping 0 kg*m/s.

    After the child throws the package, the total momentum is still going to be a whopping 0 kg*m/s. But there are two things that are contributing (or, rather, cancelling each other) to this momentum. We have the package moving some speed and we have the boat moving some speed in the other direction. Each individually has a momentum given by p = mv, but the sum of their momentums must equal zero. So that gives us the equation

    [tex]0 = p_1 + p_2 = m_1v_1 + m_2v_2[/tex]

    In which there is only a single unknown. Solve for that unknown.

  9. Apr 28, 2004 #8
    Durn...always running in a circle...at least that makes a lot of dust...

    I thought this was the same sort of problem as one I posted way back, one with people on ice skates and one pushes the other, and we have to find the velocity of the pusher, knowing the velocity of the pushee and knowing their weights...

    Well, poo. I will never really understand any of it, I guess.

    (sniffs sadly)
  10. May 4, 2004 #9
    By the way, you don't merely sum momemta. You vector sum momenta. There is a big difference between the two.
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