The impact of asteroid

In classical mechanics, linear momentum is the product of an object's mass and velocity. In a closed system (one that does not exchange any matter with the outside world and is not acted on by any outside forces) the total momentum is constant. If a huge asteroid impacts the earth in the same direction as the earth's movement, how will it change the speed of the earth? (Assume that the impact of the asteroid is a perfectly inelastic collision, the mass of the asteroid is 1 1 % of the earth's mass, and the speed of the asteroid just before the collision is 10 10 times faster than that of the earth.)

increase about 50 % will not change increase about 1.1 % increase about 8.9 %

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1 solution

The mass of the asteroid is 1 A 1A , and its speed is 10 E 10E (using the made-up units A and E referring to the mass of the asteroid and the speed of the Earth). In this unit scheme, the mass of the Earth is 100 A 100A and its speed is 1 E 1E .

Since momentum is found through the formula p = m v p=mv , the momentum of the asteroid is 10 A E 10AE and the momentum of the Earth is 100 A E 100AE . The total momentum after the collision, then, is equal to the sum of the momentums of the Earth and the asteroid, or 110 A E 110AE .

Now that we have the momentum, all that's left is the find the velocity. Rearranging p = m v p=mv gives us v = p m v=\frac{p}{m} . The total mass after the collision is 101 A 101A (100 from the Earth and 1 from the asteroid), meaning that the ending velocity is 110 A E 101 A 1.089 E \frac{110AE}{101A} \approx 1.089E . This is 8.9 % \boxed{8.9\%} greater than the starting velocity of the Earth.

i thought the answer was 1.089. how did you conclude that it's increase 8.9 percent?

Hafizh Ahsan Permana - 7 years, 2 months ago

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Same here, I got 1.089, but I realized that you have to multiply that number by 100 to get percent and then subbract 100 from it to see that its a 8.9 in percent change.

Revanth Gumpu - 6 years, 2 months ago

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