A rocket cruises past a laboratory at 0.750×106m/s in the positive x-direction j

Question

A rocket cruises past a laboratory at 0.750×106m/s in the positive x-direction just as a proton is launched with velocity (in the laboratory frame) v =(1.69×106i^+1.69×106j^)m/s.

Part A

What is the proton's speed in the laboratory frame?

Express your answer with the appropriate units.

Part B

What is the angle from the y-axis of the proton's speed in the laboratory frame?

Express your answer with the appropriate units.

Part C

What is the the proton's speed in the rocket frame?

Express your answer with the appropriate units.

Explanation / Answer

a)

The proton's speed in the laborarotary frame is given by

    v =Sqrt((1.69*106)2+((1.69*106)2 =2.390*106m/s

b)

The angle from the y-axis of the proton's speed in the laboratory frame is

Theta =tan-1(vx/vy)=tan-1((1.69*106)/(1.69*106))=45degrees

c)

The velocity of the proton in the rocket frame is

vPR =vP-vR =((1.69*106x+1.69*106y)-(0.750×106m/sx) =0.94*106x+1.69*106y

Now the speed (v) =Sqrt((0.94*106)2+(1.69*106)2) =1.9338*106m/s

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