Conservation of Momentum: Solving For Velocity After Collision

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The discussion focuses on a physics problem involving the conservation of momentum during a collision between two pucks of different masses. One puck has a momentum of mv moving right, while the other, three times as massive, moves left at -3mv. After they collide and stick together, the total mass becomes 4m, and the new velocity is calculated to be -v/2, indicating movement to the left at half the original speed. The confusion arises from understanding how momentum is conserved, as the initial total momentum is -2mv, which must equal the final momentum of the combined mass. Ultimately, the key takeaway is that momentum conservation dictates the new speed must be divided by the total mass, resulting in the final velocity being half of the original speed.
hemmi
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I'm currently reading Understanding Physics by Asimov and am stuck on an example he gives regarding conservation of momentum.

Suppose one puck was moving to the right at a given speed and had a momentum of mv, while another, three times as massive, was moving at the same speed to the left and had, therefore, a speed of -3mv. If the two stuck together after a head-on collision, the combined pucks (with a total mass of 4m) would continue moving to the left - the direction in which the more massive puck had been moving - but at half the original velocity (-v/2).

What I don't understand is, why half the velocity? -2v makes sense to me if the momentum of the small mass "canceled out" an equal momentum of the larger mass, but -1.5v doesn't make sense. I'm obviously missing something, I just don't know what. Thanks!
 
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hemmi said:
I'm currently reading Understanding Physics by Asimov and am stuck on an example he gives regarding conservation of momentum.



What I don't understand is, why half the velocity? -2v makes sense to me if the momentum of the small mass "canceled out" an equal momentum of the larger mass, but -1.5v doesn't make sense. I'm obviously missing something, I just don't know what. Thanks!


If the new velocity is -2v, then the new momentum is (4m)(-2v)=-8mv,right? But how could it have more momentum than initially given unless an external force acts?

But if no external force acts, then the momentum must be the same as it was initially given.
Initial momentum is -2mv. So if the new mass is 4m, then the new velocity must be -v/2 so that the new momentum is (4m)(-v/2)=-2mv.
 
Momentum is a vector so it can be + or - depending on direction.
p=mv m is mass in Kg and v is velocity (vector) in ms^-1 so p is momentum in Kgms^-1.
If no units are given then its just the same except it would be something like u ms^-1.
p = mv = 1*3 = 3
p = mv = 3*-3 = -9 3+-9 = -6
v = -6/4 (add up momentum and masses) = -1.5ms^-1
I think this is correct. The thing to rember is that momentum is always conserved.
 
hemmi said:
What I don't understand is, why half the velocity? -2v makes sense to me if the momentum of the small mass "canceled out" an equal momentum of the larger mass,
The momentum of the small mass does "cancel out" an equal momentum (in the opposite direction, of course) of the larger mass. The total momentum is now -2mv. To find the new speed you must divide by the mass of the entire system, which is now 4m. So: Vf = (-2mv)/4m = -v/2, in other words: half the original speed v.
 
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