Part A Two astronauts on opposite ends of a spaceship are comparing lunches. One

Question

Part A Two astronauts on opposite ends of a spaceship are comparing lunches. One has an apple, the other has an orange. They decide to trade Astronaut 1 tosses the 0.110 kg apple toward astronaut 2 with a speed of 1.14 m/s. The 0.170 kg orange is tossed from astronaut 2 to astronaut 1 with a speed of 1.20 m/s Unfortunately, the fruits collide, sending the orange off with a speed of 0.875 m/s in the negative y direction What are the final speed and direction of the apple in this case? 1,f m/s Before Submit Request Answer Part B After Ap ocounterclockwise from the positive x axis Submit Request Answer

Explanation / Answer

A)
Since there was no y momentum (Ly) originally then it must stay at 0 so

L(orange) = 0.170*(-0.875) = -0.14875 kg-m/s

now Ly(apple) = 0.14875 = 0.110*vy

==> vy = 0.14875/0.110 = 1.35m/s (+y)

The initial momentum x momentum = 0.170*1.20 - 0.110*1.14 = 0.0786
(Assuming the + x direction is the initial direction of the orange)

The final x momentum = 0.0786 = 0.110*v1x => v1x = 0.0786/.0110 =
0.715m/s (in the direction of the orange originally) (i.e the apple is
going backwards)

So v apple = sqrt(.715^2 + 1.35^2) = 1.53m/s

B)
the angle arctan(vy/vx) = arctan(1.35/0.715) = 62.09 degrees

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