Asteroid A1, mA1 3.60 * 10^6 kg, and asteroid A2, mA2 1.20 * 10^6 kg, collide he

Question

Asteroid A1, mA1 3.60 * 10^6 kg, and asteroid A2,
mA2 1.20 * 10^6 kg, collide head-on in space. Approximate
(rather poorly) the collision as being elastic. Observerswatch the event from two space platforms. An observer on
platform Q measures the speed of A1 to be 528 m>s and the
speed of A2 to be 315 m>s, and sees A2 heading directly toward
A1. The second platform, platform Z, is at rest in the
zero-momentum reference frame of the asteroids. (a) What
is the combined momentum of the asteroids measured from
platform Q? (b) What is the velocity of platform Z relative
to platform Q?

answer is .74a) 1.52×109 kg m/s 6.74b) 317 m/s in the same direction as A1

Explanation / Answer

By applying law of consevation of momentum

Pi=Pf

m1v1+m2v2=(m1+m2)v

Combined momentum measured from platform Q=m1v1+m2v2

=3.6*10^{6}*528+1.2*10^{6}*(-315)=1.52*10^{9} kg m/s

b)Velocity of Platform Z relative to Q=Vq-Vz=P/m -0 P=combined momentum and m=m1+m2

=(1.52*10^{9})/4.8*10^6 = 317 m/sec in the same direction as A1

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