Calculate the total momentum after the collision

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SUMMARY

The discussion focuses on calculating the total momentum after a collision involving two objects with equal masses of 0.55 kg. The initial velocities are 0.79 m/s East for the first object and 0 m/s for the second. After the collision, the velocities change to 0.366 m/s at 51 degrees North of East for the first object and 0.61 m/s at 21 degrees South of East for the second. The correct approach involves using vector addition to determine the momentum components in both the x and y directions, ultimately leading to the total momentum calculation.

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daewoo
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Hey guys, I'm stuck on this momentum question, i think I have an idea but have no way in proving it.


(here is part c to the question)
Calculate the total momentum after the collision. Compare this value to the total momentum before the collision. If these values differ, some likely causes of error were?

Mass(1)= 0.55kg, Mass(2)=0.55Kg, V1 =(0.79m/s East), V2= (0m/s)
V1 prime = (0.366m/s, 51 degrees N of E), and V2 Prime = (0.61m/s, 21 degrees S of E)

the prime means the change in velocity after a "collision" i first thought of using (m1v1 + m2v2) =(m1v1(prime) + m2v2(prime)) but the degrees and direction are confusing me, Unless i should use vector addition to add the angles.

Any ideas?
 
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Momentum is a vector quantity, so yes, Vector addition.
 
thought so, but how would I start it off? would i find the momentums of the primes then draw it? finding the x and y componets?
 
Yes.

Info
|\vec{v}_{1}|= 0.366 m/s
|\vec{v}_{2}|= 0.61 m/s
\theta_{1} = 51^{o}
\theta_{2} = 21^{o}

\vec{v}_{1} = |\vec{v}_{1}| \cos \theta_{1} \vec{i} + |\vec{v}_{1}| \sin \theta_{1} \vec{j}

\vec{v}_{2} = |\vec{v}_{2}| \cos \theta_{2} \vec{i} - |\vec{v}_{2}| \sin \theta_{2} \vec{j}

\vec{v}_{R} = \vec{v}_{1} + \vec{v}_{2}

\vec{p}_{total} = m \vec{v}_{R}
 
Last edited:

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