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BARSIC [14]
3 years ago
6

Two 100 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car B at –10 m/s when t

hey collide head–on. If the resulting velocity of Car B after the collision is 8 m/s, what is the velocity of Car A after the collision?
(I saw this question was asked previously, but the person did not provide the correct answer...)
Physics
1 answer:
beks73 [17]3 years ago
3 0

Answer:

v_a = -10 m/s

so car A will move with speed 10 m/s in opposite direction

Explanation:

As we know that when two cars collide then the momentum of two cars will remains conserved

so here we have

P_i = P_f

mass of two cars = 100 kg

speed of car A = 8 m/s

speed of car B = - 10 m/s

after collision the speed of car B = +8 m/s

now by momentum conservation equation

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

so we have

100(8) + 100(-10) = 100(v_a) + 100(8)

so we have

v_a = -10 m/s

so car A will move with speed 10 m/s in opposite direction

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<span>We can answer this using the rotational version of the kinematic equations:</span><span>
θ = θ₀ + ω₀<span>t + ½αt²     -----> 1</span></span>

ω² = ω₀² + 2αθ            -----> 2

Where:

θ = final angular displacement = 70.4 rad

θ₀ = initial angular displacement = 0

ω₀ = initial angular speed

ω = final angular speed

t = time = 3.80 s

α = angular acceleration = -5.20 rad/s^2

Substituting the values into equation 1:<span>
70.4 = 0 + ω₀(3.80) + ½(-5.20)(3.80)² </span><span>

ω₀ = (70.4 + 37.544) / 3.80 </span><span>

ω₀ = 28.406 rad/s </span><span>


Using equation 2:
ω² = (28.406)² + 2(-5.2)70.4 


ω = 8.65 rad/s 


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Eli rolled a ball down the cement sidewalk as hard as he could. The ball rolled for less than a minute before it came to a stop.
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According to Kepler's third law, the square of the period of revolution of planets is proportional to the cube of their mean distances from the sun. From this; T^2 = r^3.

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attashe74 [19]

Answer:

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