Conservation of Momentum Problems Draw a \"before\" and an \"after\" picture for

Question

Conservation of Momentum Problems Draw a "before" and an "after" picture for each of these situations below. In all these cases there is no net external force on the system so the total system momentum is conserved. This means that you can use the fact that the total system momentum has to be the same after an interaction as it was before to find the "unknown" quantity in the question. Example: Cart A has a mass of 3.00 kg and is going to the right at a speed of 0.200 m mass of 2.00 kg and is going to the left with a speed of 0.300 m/s. They collide and cart A is seen to now be going to the left with a speed of 0.100 m/s. What is the velocity of B after the collision? /s and cart B has a 0.200 mis 0.300 m/s v = ?? 0.100 m/s does not change AS PRE- 0.3 2kg : 0.1 5m 0.15 0306 0.1s 1. Now use the example as a model and do the following problems the same way: The same two carts as above have the same initial velocities as the example above, but when they hit they stick together, so that the final situation is essentially a big 5.00 kg cart with some final momentum and velocity that you can find a.

Explanation / Answer

a) Applying conservation of momentum

=>    3 * 0.200 - 2 * 0.300 = (3 + 2) Vf

=> Final velocity = 0 m/sec

=>   Final momentum = 0 kg-m/sec

b)    Applying conservation of momentum

=>    3 * 0.500 - 2 * 0.100 = (3 + 2) Vf

=> Final velocity = 0.26 m/sec

=>   Final momentum = 1.3 kg-m/sec

c)    Applying conservation of momentum

=>    5 * 0.200 = - 3 * 0.100 + 2 * Vf

=> speed of 2 kg cart = 0.65   m/sec

d)      Ist case

    average force on cart A =   0 - (3 * 0.200)/0.100

                                           =    - 6 N

   2nd case

    average force on cart A =   ((3 * 0.26) - (3 * 0.500))/0.100

                                           =    - 7.2 N

3rd case

    average force on cart A =   ((- 3 * 0.100) - (3 * 0.200))/0.100

                                           =    - 9 N

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