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4 years ago

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A(n) 1.51 kg sphere makes a perfectly inelastic collision with a second sphere that is initially at rest. The composite system m

oves with a speed equal to one-third the original speed of the 1.51 kg sphere. What is the mass of the second sphere?

1 answer:

marusya05 [52]4 years ago

6 0

Answer: 3.02kg

Explanation:

According to the law of conservation of momentum, the sum of momentum of bodies before collision is equal to the sum of momentum of the bodies after collision.

Note that this bodies will move with a common velocity after collision.

Since momentum = mass of a body × its velocity

Let m1 and m2 be the masses of the spheres

u1 and u2 be their velocities

v be their common velocity after collision

Mathematically

m1u1 + m2u2 = (m1+m2)v

From the question, the second sphere is initially at rest i.e u2 = 0m/s and the composite system moves with a speed equal to one-third the original speed of 1st sphere i.e V = u1/3

Substituting this conditions into the formula, we have;

m1u1 + m2(0) = (m1+m2)u1/3

m1u1 = u1/3(m1+m2)

Given m1 = 1.51kg

m1 = 1/3(m1+m2)

m1 = m1/3 + m2/3

m1-m1/3 = m2/3

2m1/3 = m2/3

2m1 = m2

2(1.51) =m2

m2 = 3.02kg

The mass of the second sphere is 3.02kg

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Explanation:

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From the free body diagram we can build an equation with the sum of forces between the start and the planet.

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Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

$F_{c}=ma_{c}$

where the centripetal acceleration is given by:

$a_{c}=\omega ^{2}r$

where

$\omega = \frac{2\pi}{T}$

Where T is the period, and $\omega$ is the angular speed of the planet, so:

$a_{c} = ( \frac{2\pi}{T})^{2}r$

or:

$a_{c}=\frac{4\pi^{2}r}{T^{2}}$

so:

$F_{c}=m(\frac{4\pi^{2}r}{T^{2}})$

so now we can do the sum of forces:

$\sum F=ma$

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in this case we can get rid of the mass of the planet, so we get:

$G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})$

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$M_{2}$ = mass of Star 2

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$\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}$

we can get rid of all the constants so we end up with:

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$M_{2}>M_{1}$

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