Part A nstant Two astronauts on opposite ends of a spaceship are comparing lunch

Question

Part A nstant 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 vi1 1.13 m/s. The 0.170 kg orange is tossed from astronaut 2 to astronaut 1 with a speed of 1.25 m/s Unfortunately, the fruits collide, sending the orange off with a speed of 1.01 m/s in the negative y direction What are the final speed and direction of the apple in this case? vE,2- 1.41 m/s Previous Ans Completed Before Part B After o counterclockwise from the positive x axis Submit Provide Feedback

Explanation / Answer

Using law of conservation of momentum:

Before:

x direction: 0.130*1.13 - 0.170 * 1.25 = - 0.0656 Kg * m / s

y direction = 0

After:

Vay is velocity of the apple in the y direction, Vax is the velocity of the apple in the x direction

momentum again:

x direction: 0.130 * Vax + 0 = -0.0656 (momentum did not change) hence: Vax = -0.50 m/s

y direction: 0.130 * Vay -0.170 * 1.01 = 0 hence Vay = 1.32 m/s

so the apple has a speed of sqrt( (-0.50)^2 + 1.32^2) = 1.41 m/s

and the angle is in the second quadrant using tg(angle) = (Vay)/(Vax) = 110.75 degrees.

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