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3 years ago

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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...)

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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A stone is thrown vertically upwards with an initial velocity of 20m/sec. Find the maximum height ot reaches and the time taken

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Answer:

The height reached is 20m, The time taken to reach 20m is 2 seconds

Explanation:

Observing the equations of motion we can see that the following equation will be most helpful for this question.

$v^{2} = u^{2} + 2as$

We are given initial velocity, u

We know that the stone will stop at its maximum height, so final velocity, v

Acceleration, a

And we are looking for the displacement (height reached), s

Substitute the values we are given into the equation

$0^{2} = 20^{2} + 2(10)s$

Rearrange for s

$0^{2} -20^{2} =20s$

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$\frac{-400}{20} =s$

s = -20 (The negative is just showing direction, it can be ignored for now)

The height reached is 20m

Use a different equation to find the time taken

$s = vt - \frac{1}{2} at^{2}$

Substitute in the values we have

$-20=(0)t - \frac{1}{2} (10)t^{2}$

Rearrange for t

$-20 =0 -5 t^{2}$

$\frac{-20}{-5} =t^{2}$

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By the formula of magnification

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Answer:

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When calculating the electric flux, the angle between the directions of electric field and the area becomes important, especially if the angle is changing with time.

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