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3 years ago

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A particle moving at 10 m/s along the x-axis collides elastically with another particle moving at 5.0 m/s in the same direction

along the x-axis. the particles have equal masses. what are their speeds after the collision? Can someone please explain this in the simplest way possible?

1 answer:

drek231 [11]3 years ago

8 0

Explanation:

It is given that,

Velocity of particle 1, u₁ = 10 m/s

Velocity of particle 2, u₂ = 5 m/s

Let v₁ and v₂ are the final speed of both particles after the collision. Applying the conservation of momentum as :

$mu_1+mu_2=mv_1+mv_2$

$15=v_1+v_2$ ......................(1)

For an elastic collision, the coefficient of restitution is equal to 1 as :

$\dfrac{v_2-v_1}{u_1-u_2}=1$

$\dfrac{v_2-v_1}{5}=1$

${v_2-v_1}=5$ ................(2)

On solving equation (1) and (2), we get,

$v_1=5\ m/s$

$v_2=10\ m/s$

So, the speeds of particle 1 and particle 2 after the collision is 5 m/s and 10 m/s respectively. Hence, this is the required solution.

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Answer:

μsmín = 0.1

Explanation:

There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
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This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
This force has the following general expression:

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Since Fc is actually Fn, we can replace the right side of (2) in (1), as

follows:

$F_{frmax} = m* \mu_{s} * \omega^{2} * r (3)$

When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

$m* g = m* \mu_{smin} * \omega^{2} * r (4)$

(The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)

Cancelling the masses on both sides of (4), we get:

$g = \mu_{smin} * \omega^{2} * r (5)$

Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

$60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)$

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A 1000 kg car moving a 10 m/s collides with a stationary 2000 kg truck. The two vehicles interlock as a result of the collision.

IgorLugansk [536]

Answer:

v₃ = 3.33 [m/s]

Explanation:

This problem can be easily solved using the principle of linear momentum conservation. Which tells us that momentum is preserved before and after the collision.

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