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shutvik [7]
3 years ago
11

In an isolated system, bicycle 1 and bicycle 2, each with a mass of 15 kg, collide. bicycle 1 was moving to the right at 1 m/s,

while bicycle 2 was moving to the left at 3 m/s. what is the magnitude of their combined momentum before the crash?
Physics
1 answer:
IgorC [24]3 years ago
6 0
To do this, all you need to do is get the momentum of each and get the sum. The formula of momentum is:

p=mv

where:
p = momentum
m = mass
v = velocity

Bicycle 1 and 2 have the same mass of 15 kg

Bicycle 1 v = 1 m/s
Bicycle 2 v = 3 m/s but because it was coming from the opposite direction then we get a negative velocity. The negative just indicates the direction so it will be -3m/s

m1v1+m2v2

=(15kg)(1m/s) + (15kg)(-3m/s) 
= 15kg.m/s + (-45kg.m/s)
=-30kg.m/s

Their combined momentum before the crash is 30kg.m/s. We take out the negative sign because the question is asking only for the magnitude and not the direction and remember the negative sign indicates the direction. 

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The tralational equilibrium condition allows finding that the electric potential is   V = 4.8 10¹¹ V

Given parameter

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Newton's second law states that the force is proportional to the mass and the acceleration of the bodies, in the special case that the acceleration is zero, it establishes the condition for the equilibrium of the bodies

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x- axis

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It's solve the system of equations

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It's calculate

         V = \frac{5.00 \ 10^{-2} 1.5 \ 10^{-3} \ 9.8}{ 8.9 10^{-16} } \ tan \ 30

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In conclusion, using the equilibrium condition, we could find that the electric potential is V = 4.8 10¹¹ V

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