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umka21 [38]
3 years ago
13

Two identical particles of charge 6 μμC and mass 3 μμg are initially at rest and held 3 cm apart. How fast will the particles mo

ve when they are allowed to repel and separate to very large (essentially infinite) distance? Now suppose that the two particles have the same charges from the previous problem, but their masses are different. One particle has mass 3 μμg as before, but the other one is heavier, with a mass of 30 μμg. Their initial separation is the same as before. How fast are the particles moving when they are very far apart?
Physics
1 answer:
krek1111 [17]3 years ago
3 0

Answer:

Explanation:

The charges will repel each other and go away with increasing velocity , their kinetic energy coming from their potential energy .

Their potential energy at distance d

= kq₁q₂ / d

= 9 x 10⁹ x 36 x 10⁻¹² / 2 x 10⁻² J

= 16.2 J

Their total kinetic energy will be equal to this potential energy.

2 x 1/2 x mv² = 16.2

= 3 x 10⁻⁶ v² = 16.2

v = 5.4 x 10⁶

v = 2.32 x 10³ m/s

When masses are different , total P.E, will be divided between them as follows

K E of 3 μ = (16.2 / 30+3) x 30

= 14.73 J

1/2 X 3 X 10⁻⁶ v₁² = 14.73

v₁ = 3.13 x 10³

K E of 30 μ = (16.2 / 30+3) x 3

= 1.47 J

1/2 x 30 x 10⁻⁶ x v₂² = 1.47

v₂ = .313 x 10³ m/s

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a) \omega=\frac{-mvR}{I+MR^2}

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Explanation:

a)

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At the beginning, the merry-go-round and the girl are at rest, so the initial angular momentum is zero:

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L_r=mvR is the angular momentum of the rock, where

m is the mass of the rock

v is its velocity

Since the total angular momentum is conserved,

L_1=L_2

So we find:

0=(I+I_g)\omega +mvR\\\omega=\frac{-mvR}{I+MR^2}

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b)

The linear speed of a body in rotational motion is given by

v=\omega r

where

\omega is the angular speed

r is the distance of the body from the axis of rotation

In this problem, for the girl, we have:

\omega=\frac{-mvR}{I+MR^2} is the angular speed

r=R is the distance of the girl from the axis of rotation

Therefore, her linear speed is:

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Answer:

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From the law of conservation of energy, U₁ + K₁ = U₂ + K₂

So, kQ²/d + 0 = kQ²/3d + K

K₂ = kQ²/d - kQ²/3d = 2kQ²/3d

So, the kinetic energy K₂ of the moving charge is K₂ = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd

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