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Elastic and Inelastic collisions assignement problems. 1st year physics

  1. Nov 5, 2011 #1
    Here are 3 problems I have been working on. Any help would be very much appreciated. Thank you in advance!



    1. The problem statement, all variables and given/known data
    Two inelastic balls are traveling toward eachother with velocities -4m/s and 7 m/s and they experience an elastic head on collision
    a)obtain the velocities (magnitude and direction) of each ball after the collision
    b) recalculate the results for the ball moving at 7m/s being 2x the mass of the other ball


    2. Relevant equations
    m1v1 + m2v2 = m1V1 + m2V2
    v1 + V1 + v2 + V2

    where v= initial and V=final

    3. The attempt at a solution

    i feel to solve this i need to substitute one of the equations into the other and solve for one of the final velocities.

    I did this and obtained

    m1v1 + m2v2 - m1v2 + m1v1 = V (m1 +m2)

    since masses are equal i factored like this

    m(v1 +v2 -v2 +v1)/2m = V
    m(v1+v1)/2m = V
    mv1(2)/2m = V
    v1=V

    I feel like this is wrong. I have done it a few times and get different answers. I dont account for the direction of ball 2 until i plug in a value as -4, until then I leave it as positive.





    1. The problem statement, all variables and given/known data

    A 0.5kg mass of silly putty is at rest at the top of a 100m tall building; it is dropped. At the same time a 0.003kg paintball pellet is fight straight up towards the ground with an initial veloctiy of 150 m/s. The paintball and putty collide inelastically 2.18m below the top of the building. Measured from the ground what is the maximun height that the combined putty/paintball mass achieves?

    2. Relevant equations

    v^2 = v0^2 +2ax

    m1v01 + m2v02 = (m1 +m2)Vf

    3. The attempt at a solution

    i first used kinematics to find the velocities of the objects right at impact. the putty = -6.54m/s and the pellet =143.46.

    Next I used m1v01 + m2v02 = (m1 +m2)Vf to solve for the velocity of the combined mass right after the collision. I found it to be -5.65m/s. I am not sure why my answer is negative and cant see why but I know it should be a positive value. If i use the positive value and use kinematics to solve for the height once it decelerates to v=0m/s i find it goes 1.63m equaling 99.45m above the ground.

    I am unsure of this answer. I feel like my method is right but I am not sure if what I did was correct.


    1. The problem statement, all variables and given/known data
    1 0.5kg mass of silly putty in suspended 5m above the floor by a long string. A paint ball pellet (m=0.003kg) is fired at the putty from the floor with an initial velocity of 150m/s. the two masses collide inelastically. what is the minimum length of the string such that the combined putty/paint ball mass does not splatter into the ceiling?


    2. Relevant equations

    v^2 = v0^2 + 2ax
    m1v01 +m2v02 = )m1+m2)Vf



    3. The attempt at a solution

    using kinematics i determined the speed of the pellet before impact would be 149.67m/s. Using the equation for inelastic collision i found that the v0 of the ball is 0 and the vf of the combined mass to be 0.89m/s. Using this value I used the kinematics equation again and found distance (x) it would go until it came to rest. I found this value to be 0.04m.

    After working through this problem I dont see what I am doing wrong. I dont think my answer makes sense since the string would only have to be longer than 4cm.
     
  2. jcsd
  3. Nov 6, 2011 #2

    PeterO

    User Avatar
    Homework Helper

    Your answer to part 3 is correct. The paint ball pellet has almost negligible mass compared to the 0.5 kg mass, so the pair will move up very little.

    In Part 2, your answer of -5.65 is correct. The falling mass will be slowed slightly, but not stopped, so the maximum height reahed by the pair is the point of impact.

    For Q1, try calculating the velocity of the centre of mass. Each ball will be approaching the centre of mass at a certain speed. They will be leaving the centre of mass at those same speeds after collision. Even when the masses are different.
     
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