Two asteroids strike head-on: before the collision, asteroid A (mA = 8.0×1012 kg

Question

Two asteroids strike head-on: before the collision, asteroid A (mA = 8.0×1012 kg ) has velocity 3.3 km/s and asteroid B (mB = 1.28×1013 kg ) has velocity 1.4 km/s in the opposite direction. If the asteroids stick together, what is the magnitude of the velocity of the new asteroid after the collision? If the asteroids stick together, what is the direction of the velocity of the new asteroid after the collision? If the asteroids stick together, what is the direction of the velocity of the new asteroid after the collision? in the original direction of asteroid B, or in the original direction of asteroid A

Explanation / Answer

Apply conservation of momentum to the collision
Taking the initial direction of A as +ve

before ..
(8.0*1012kg x 3.30km/s) + (1.28*1013kg x -1.40km/s) = 8.48*1012 kg.km/s
after (joined) ..
(8*1012 + 1.28*1013)kg x Vf = (2.08*1013 x Vf) kg.km/s

before = after ..
8.48*1012 = 2.08*1013 x Vf
Vf = (+) 0.4077 km/s

Vf turns-out +ve .. it's in the same original direction as A.

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