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Marina CMI [18]
3 years ago
6

The following diagrams show five pairs of asteroids, labeled with their relative masses (M) and distances (d) between them. For

example, an asteroid with M=2 has twice the mass of one with M=1 and a distance of d=2 is twice as large as a distance of d=1. Rank each pair from left to right based on the strength of the gravitational force attracting the asteroids to each other, from strongest to weakest.
Physics
1 answer:
serg [7]3 years ago
7 0

Answer:

- d = 2; m = 2; m = 2

- d= 1; m = 1; m = 2

- d = 1; m = 1; m = 1

- d = 2; m = 1; m = 2

- d = 2; m = 1; m = 1

Explanation:

The following diagrams show five pairs of asteroids, labeled with their relative masses (M) and distances (d) between them. For example, an asteroid with M=2 has twice the mass of one with M=1 and a distance of d=2 is twice as large as a distance of d=1. Rank each pair from left to right based on the strength of the gravitational force attracting the asteroids to each other, from strongest to weakest.

- d = 2; m = 2; m = 2

- d= 1; m = 1; m = 2

- d = 1; m = 1; m = 1

- d = 2; m = 1; m = 2

- d = 2; m = 1; m = 1

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Block 1, of mass m₁ = 1.30 kg , moves along a frictionless air track with speed v₁ = 29.0 m/s. It collides with block 2, of mass
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a. 37.7 kgm/s b. 0.94 m/s c. -528.85 J

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a. The initial momentum of block 1 of m₁ = 1.30 kg with speed v₁ = 29.0 m/s is p₁ = m₁v₁ = 1.30 kg × 29.0 m/s = 37.7 kgm/s

The initial momentum of block 2 of m₁ = 39.0 kg with speed v₂ = 0 m/s since it is initially at rest is p₁ = m₁v₁ = 39.0 kg × 0 m/s = 0 kgm/s

So, the magnitude of the total initial momentum of the two-block system = (37.7 + 0) kgm/s = 37.7 kgm/s

b. Since the blocks stick together after the collision, their final momentum is p₂ = (m₁ + m₂)v where v is the final speed of the two-block system.

p₂ = (1.3 + 39.0)v = 40.3v

From the principle of conservation of momentum,

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37.7 kgm/s = 40.3v

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So the final velocity of the two-block system is 0.94 m/s

c. The change in kinetic energy of the two-block system is ΔK = K₂ - K₁ where K₂ = final kinetic energy of the two-block system = 1/2(m₁ + m₂)v² and K₁ = final kinetic energy of the two-block system = 1/2m₁v₁²

So, ΔK = K₂ - K₁ = 1/2(m₁ + m₂)v² - 1/2m₁v₁² = 1/2(1.3 + 39.0) × 0.94² - 1/2 × 1.3 × 29.0² = 17.805 J - 546.65 J = -528.845 J ≅ -528.85 J

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