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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 0.132 kg apple toward astronaut 2 with a speed of 1.25 m/s. The 0.143 kg orange is tossed from astronaut 2 to astronaut 1 with a speed of 1.14 m/s. Unfortunately, fruits collide, sending the orange off with a speed 1.03 m/s and an angle of 43.0° with respect to its original direction of motion. Using conservation of linear momentum, find the final speed and direction of the apple. Assume an elastic collision occurs. Give the apple’s direction relative to its original direction of motion.

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Answer:

Step-by-step explanation:

We shall consider direction towards left as positive Let the required velocity be v and let v makes an angle φ

Applying law of conservation of momentum along direction of original motion

m₁ v₁ - m₂ v₂ = m₂v₃ - m₁ v₄

0.132 x 1.25 - .143 x 1.14 = 1.03 cos43 x .143 - v cos θ

v cos θ = .8

Applying law of conservation of momentum along direction perpendicular to direction of original motion

1.03 sin 43 x .143 = .132 x v sinθ

v sinθ = .76

squaring and adding

v² = .76 ² + .8²

v = 1.1 m /s

Tan θ = .76 / .8

θ = 44°

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