PHY 201
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A steel ball of mass 60 grams, moving at 80 cm / sec, collides with a stationary marble of mass 20 grams. As a result of the collision the steel ball slows to 50 cm / sec and the marble speeds up to 70 cm / sec.
• Is the total momentum of the system after collision the same as the total momentum before?
answer/question/discussion: ->->->->->->->->->->->-> :
The total momentum before the collision is: .06 kg * .8 m/s + .02 kg * 0 m/s = .048 N. The total momentum after the collision is: .06 kg * .5 m/s + .02 kg * .7 m/s = .044 N. There are not exactly the same, so no, they are not equal.
The Newton is the unit mass * acceleration, i.e., kg * m/s^2. The unit here is kg * m/s. This unit is not the Newton.
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• What would the marble velocity have to be in order to exactly conserve momentum, assuming the steel ball's velocities to be accurate?
answer/question/discussion: ->->->->->->->->->->->-> :
The momentum after the collision is all that we are concerned about, because the marble at the beginning is 0 m/s, so we solve for the velocity of the marble after the collision with the total momentum before the collision (m = marble):
.06 kg * .5 m/s + .02 kg * vf_m = .048 N.
.03 N + .02 kg * vf_m = .048 N
.05 N kg * vf_m = .048 N Divide the .05 N kg to both sides
vf_m = .96 m/s
You have a good equation.
However .03 N + .02 kg * vf_m cannot be simplified. One is in units of Newtons (should actually be kg m/s) and the other is in kg * vf_m. These can't be combined into a singe term any more that 5 x^2 could be combined with 7 y.
So, the velocity of the marble would have to be .96 m/s, in order for the momentums before and after the collision to be equal.
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30 minutes
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