To solve this problem, we need to use the principle of conservation of momentum. The momentum before the collision is equal to the momentum after the collision.

The momentum of the meteor before the collision is its mass times its velocity, which is (5 x 10^10 kg) * (7200 m/s) = 3.6 x 10^14 kg*m/s.

The Earth's mass is approximately 5.972 x 10^24 kg. After the collision, the Earth and the meteor move together, so their combined mass is the Earth's mass plus the meteor's mass, which is approximately still the Earth's mass since the meteor's mass is negligible compared to the Earth's mass.

So, we can set up the equation for the conservation of momentum as follows:

(mass of meteor * velocity of meteor) = (mass of Earth * velocity of Earth after collision)

Solving for the velocity of the Earth after the collision gives us:

velocity of Earth after collision = (mass of meteor * velocity of meteor) / mass of Earth

= (3.6 x 10^14 kg*m/s) / (5.972 x 10^24 kg)

= 6.03 x 10^-11 m/s

So, the speed that such a meteor would give the Earth in a head-on collision is approximately 6.03 x 10^-11 m/s.

Question

To solve this problem, we need to use the principle of conservation of momentum. The momentum before the collision is equal to the momentum after the collision.

The momentum of the meteor before the collision is its mass times its velocity, which is (5 x 10^10 kg) * (7200 m/s) = 3.6 x 10^14 kg*m/s.

The Earth's mass is approximately 5.972 x 10^24 kg. After the collision, the Earth and the meteor move together, so their combined mass is the Earth's mass plus the meteor's mass, which is approximately still the Earth's mass since the meteor's mass is negligible compared to the Earth's mass.

So, we can set up the equation for the conservation of momentum as follows:

(mass of meteor * velocity of meteor) = (mass of Earth * velocity of Earth after collision)

Solving for the velocity of the Earth after the collision gives us:

velocity of Earth after collision = (mass of meteor * velocity of meteor) / mass of Earth

= (3.6 x 10^14 kg*m/s) / (5.972 x 10^24 kg)

= 6.03 x 10^-11 m/s

So, the speed that such a meteor would give the Earth in a head-on collision is approximately 6.03 x 10^-11 m/s.

Knowee AI · Accepted Answer

To solve this problem, we need to use the principle of conservation of momentum. The momentum before the collision is equal to the momentum after the collision.

The momentum of the meteor before the collision is its mass times its velocity, which is (5 x 10^10 kg) * (7200 m/s) = 3.6 x 10^14 kg*m/s.

The Earth's mass is approximately 5.972 x 10^24 kg. After the collision, the Earth and the meteor move together, so their combined mass is the Earth's mass plus the meteor's mass, which is approximately still the Earth's mass since the meteor's mass is negligible compared to the Earth's mass.

So, we can set up the equation for the conservation of momentum as follows:

(mass of meteor * velocity of meteor) = (mass of Earth * velocity of Earth after collision)

Solving for the velocity of the Earth after the collision gives us:

velocity of Earth after collision = (mass of meteor * velocity of meteor) / mass of Earth

= (3.6 x 10^14 kg*m/s) / (5.972 x 10^24 kg)

= 6.03 x 10^-11 m/s

So, the speed that such a meteor would give the Earth in a head-on collision is approximately 6.03 x 10^-11 m/s.

Meteor Crater in Arizona is thought to have been formed by the impact of a meteor with the Earth some 20,000 years ago. The mass of the meteor is estimated at 5 ✕ 1010 kg. Its speed is estimated at 7200 m/s. What speed would such a meteor give Earth in a head-on collision?

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