Two astronauts on opposite ends of a spaceship are comparing lunches. One has an

Question

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 0.130 kg apple toward astronaut 2 with a speed of vi,1 = 1.09 m/s . The 0.170 kg orange is tossed from astronaut 2 to astronaut 1 with a speed of 1.18 m/s . Unfortunately, the fruits collide, sending the orange off with a speed of 0.964 m/s in the negative y direction.

What are the final speed and direction of the apple in this case?

Vf,2 = ??m/s

Part B

counterclockwise from the positive x axis

Explanation / Answer

m1 = 0.13 Kg
v1i = 1.09 m/s

Assume orange is going from left to right and apple is going from right to left

m2 = 0.170 Kg
v2i = 1.18 m/s


Using momentum conservation,
In Horizontal direction,
m1 * v1i - m2 * v2i = 0.130 * v * cos( )
0.130 * v * cos() =  (0.170*1.18) - (0.130*1.09)
v * cos() = 0.453

In Vertical Direction,
0.130 * v * sin() = 0.170 * 0.964
v * sin() = 1.26

= tan^-1(1.26/0.453)
= 70.23 o

v = 1.26/sin(70.2)
v = 1.34 m/s

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