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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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George Of The Jungle's wife, Mrs. Of The Jungle, has been pestering him to go on a diet. He should have listened. During his com
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Answer:

112.06 kg - Thats heavy !

Explanation:

Let's do force balance here. Let the object of our interest be George. The forces acting on him are the tension in  the upward direction, his weight in the downward direction and the centrifugal force in the downward direction. Considering the upward and downward directions on the y-axis and f=given the fact that George doesn't move up or down, the forces are balanced along the y-axis. Hence doing force balance:

magnitude of forces upward =magnitude of forces downward

i.e., Tension(T) = Weight(mg) + Centrifugal force (mv²/r)

where: 'm' is the mass of George, g is the acceleration due to gravity (9.8 m/s²). v is the speed with which George moves (14.1 m/s) and r is the radius of the circle in which he's moving at the instant (Here since he's swinging on the rope, he moves in a circle with radius as the length of the rope and hence r=7.3m).

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7 0
3 years ago
A horizontal circular platform (m = 119.1 kg, r = 3.23m) rotates about a frictionless vertical axle. A student (m = 54.3kg) walk
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Answer:

\omega_2=5.1rad/s

Explanation:

Since there is no friction angular momentum is conserved. The formula for angular momentum thet will be useful in this case is L=I\omega. If we call 1 the situation when the student is at the rim and 2 the situation when the student is at r_2=1.39m from the center, then we have:

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Or:

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And we want to calculate:

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The total moment of inertia will be the sum of the moment of intertia of the disk of mass m_D=119.1 kg and radius r_D=3.23m, which is I_D=\frac{m_Dr_D^2}{2}, and the moment of intertia of the student of mass m_S=54.3kg at position r (which will be r_1=r=3.23m or r_2=1.39m) will be I_{S}=m_Sr_S^2, so we will have:

\omega_2=\frac{(I_D+I_{S1})\omega_1}{(I_D+I_{S2})}

or:

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6 0
3 years ago
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