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Jlenok [28]
4 years ago
12

6) Two small balls, A and B, attract each other gravitationally with a force of magnitude F. If we now double both masses and th

e separation of the balls, what will now be the magnitude of the attractive force on each one?
Physics
1 answer:
Butoxors [25]4 years ago
4 0

Answer:

The magnitude of the attractive force is unchanged

Explanation:

Newton's Law of Universal Gravitation: The law states that the force of attraction between two mass m₁ and m₂ is directly proportional to the product of the masses and inversely proportional to the square of the distance (d) between them.

Mathematically it is represented as

F = Gm₁m₂/d₂ ............................. Equation 1

Where G = universal Constant, F = Force of attraction or repulsion, m₁ = mass of the first body,  m₂ = mass of the second body, d = distance between the masses.

When both masses and the separation of the ball are doubled,

I.e

Fₙ = G(2m₁)(2m₂)/(2d)²

Where Fₙ = the new gravitational force, when both masses and distance are separation are doubled.

Fₙ = 4Gm₁m₂/4d²

Fₙ = Gm₁m₂/d²........................... Equation 2

Comparing Equation 1 and equation 2,

Fₙ = F

Therefore the magnitude of the attractive force between the balls when their masses and separation are doubled remains the same.

I.e The magnitude of the attractive force is unchanged,

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Answer:

a) t = \sqrt{\frac{h}{2g}}

b) Ball 1 has a greater speed than ball 2 when they are passing.

c) The height is the same for both balls = 3h/4.

Explanation:

a) We can find the time when the two balls meet by equating the distances as follows:

y = y_{0_{1}} + v_{0_{1}}t - \frac{1}{2}gt^{2}  

Where:

y_{0_{1}}: is the initial height = h

v_{0_{1}}: is the initial speed of ball 1 = 0 (it is dropped from rest)

y = h - \frac{1}{2}gt^{2}     (1)

Now, for ball 2 we have:

y = y_{0_{2}} + v_{0_{2}}t - \frac{1}{2}gt^{2}    

Where:

y_{0_{2}}: is the initial height of ball 2 = 0

y = v_{0_{2}}t - \frac{1}{2}gt^{2}    (2)

By equating equation (1) and (2) we have:

h - \frac{1}{2}gt^{2} = v_{0_{2}}t - \frac{1}{2}gt^{2}

t=\frac{h}{v_{0_{2}}}

Where the initial velocity of the ball 2 is:

v_{f_{2}}^{2} = v_{0_{2}}^{2} - 2gh

Since v_{f_{2}}^{2} = 0 at the maximum height (h):

v_{0_{2}} = \sqrt{2gh}

Hence, the time when they pass each other is:

t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}

b) When they are passing the speed of each one is:

For ball 1:

v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh}

The minus sign is because ball 1 is going down.

For ball 2:

v_{f_{2}} = v_{0_{2}} - gt = \sqrt{2gh} - g*\sqrt{\frac{h}{2g}} = (\sqrt{1} - \frac{1}{\sqrt{2}})*\sqrt{gh} = 0.41\sqrt{gh}

Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.

c) The height of the ball is:

For ball 1:

y_{1} = h - \frac{1}{2}gt^{2} = h - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h

For ball 2:

y_{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} = \sqrt{2gh}*\sqrt{\frac{h}{2g}} - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h

Then, when they are passing the height is the same for both = 3h/4.

I hope it helps you!                  

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