Sixty five million years ago a spherical asteroid with a uniform density if 4100

Question

Sixty five million years ago a spherical asteroid with a uniform density if 4100 kg/m^2 and a diameter of 9600 m collided with the Earth. The collision was head-on; the velocity of the asteroid before impact was 25000 m/s, and the velocity of the Earth before impact was 30000 m/s in the opposite direction. (The mass of the Earth was, and still is 5.98 x 10^ 24 kg. Also, the asteroid was destroyed during the collision but it’s entire mass remained on the Earth.) Note: density = mass x volume and the asteroid is spherical.

A) determine the change in velocity of the earth due to the asteroid collision.

B) determine the amount of kinetic energy lost during the collision. (State your answer in terms of KE (kinetic energy) and as a percentage of the original total KE of the Earth/asteroid system total KE.

Explanation / Answer

mass of asteroid = volume x density

= (4 x pi x 9600^3 / 3) (4100)

= 1.519 x 10^16 kg

(A) Applying momentum conservation,

( 5.98 x 10^24)(300000) - (1.519 x 10^16 x 250000) = (5.98 x 10^24 + 1.519 x 10^16) vf

change in velocity = 30000 - vf

= [(5.98 x 10^24 x 30,0000) + (1.519 x 10^16 x 30,000) - (5.98 x 10^24 x 30,000) + (1.519 x 10^16 x 25,000)]/(5.98 x 10^24 + 1.519 x 10^16)

= 1.40 x 10^-4 m/s

(B) lost KE = Ki - Kf

=[ mE x vE^2 /2 + mS x vS^2 /2 ] - [ (mE +mS ) ( mE vE - mS vS)/(mE + ms))^2 / 2]

= [ mE x vE^2 /2 + mS x vS^2 /2 ] - [( mE vE - mS vS)^2/2(mE + ms)]

= [ mE^2 vE^2 + mS^2 Vs^2 + mE ms VE^2 + ms mE vS^2 - mE^2 vE^2 - mS^2 vS^2 + 2 mE ms vE vS]/2(ms + mE)

= [ mE ms VE^2 + ms mE vS^2 + 2 mE ms vE vS] / 2 (ms + mE)

= 2.29 x 10^25 J  

percentage loss = {[ mE ms VE^2 + ms mE vS^2 + 2 mE ms vE vS] / 2 (ms + mE) } / { ms vE^2 / 2} x 100

= 8.50 x 10^-7 %

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.