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Ilya [14]
3 years ago
5

Two marbles (m = 0.050 kg each) moving on a table TOWARDS each other with the speed of 0.3 m/s collide, and start moving away fr

om each other.
a) What is the total momentum of this system before the collision?
V1 = __________ m/s, V2 = __________ m/s,

p1 = __________ kg m/s, p2 = __________ kg m/s,

ptotal = ____________ kg m/s
Physics
1 answer:
snow_tiger [21]3 years ago
7 0

Answer:

V1 = ____0.3______ m/s, V2 = ____-0.3______ m/s,

p1 = ____0.015______ kg m/s, p2 = ___-0.015_______ kg m/s,

ptotal = _0___________ kg m/s

Explanation:

Taking the marble ball that moves from right to left as Marble 1 and the one that moves from left to right as Marble 2. We take direction from left to right as negative and from right to left as positive. Therefore, velocity of marble 1 will be +0.3 m/s while velocity of marble 2 will be -0.3 m/s. The mass of marble 1, m2 0.05 Kg and same to marble 2, m2=0.05 Kg

Momentum is product of mass and velocity hence using p to represent momentum

P1=0.05*0.3=0.015 Kg m/s

P2=0.05*-0.3=-0.015 Kg m/s

V1=0.3 m/s, V2=-0.3 m/s

The total momentum before collision will be the sum of P1 and P2 hence 0.015 Kg m/s+-0.015 Kg m/s=0

Ptotal=0 Kg m/s

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Same direction: t=234s; d=6.175Km

Opposite direction: t=27.53s; d=0.73Km

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If the automobile and the train are traveling in the same direction, then the automobile speed relative to the train will be v_{AT}=v_A-v_T (<em>the train must see the car advancing at a lower speed</em>), where v_A is the speed of the automobile and v_T the speed of the train.

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t = \frac{L}{v_{AT}} = \frac{1.3Km}{170Km/h} = 0.00765h=27.53s

d=v_At=(95Km/h)(0.00765h)=0.73Km

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