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 = 1.05 m/s . The 0.150 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 1.03 m/s in the negative y direction. What are the final speed and direction of the apple in this case?

Answer 1

Answer: v= 1.23 m/s θ = 75.3º

Step-by-step explanation:

First of all, we define the direction in which both fruits are tossed as the x axis, so all initial momenta have horizontal components only.

Now, if no external forces act during collision (due to the infinitesimal time during which collision takes place) momentum must be conserved.

As momentum is a vector, both components must be conserved, so we can write the following equations:

p₁ₓ = p₂ₓ ⇒ -m₁ . vi₁ +m₂. vi₂ = m₁ . vf₁ . cos θ (1)

p₁y = p₂y ⇒ 0 =m₂ . vf₂ - m₁. vf₁. sin θ (2)

Replacing by the values of m1, m₂, vi₁, vi₂, and vf₂, we can calculate the value of the angle θ, that the apple forms with the horizontal, as follows:

(1) -0.13 Kg. 1.05 m/s + 0.15 Kg. 1.18 m/s = 0.13. vf . cos θ

(2) 0.15 Kg. 1.03 m/s = 0.13 vf. sin θ

sin θ / cos θ = 3.82 ⇒ tg θ = 3.82 ⇒ θ = arc tg (3.82) = 75.3º

Replacing this value of θ in (2), we get:

0.15 kg. 1.03 m/s = 0.13 vf . sin 75.3º = 0.13 . vf . 0.967

Solving for vf, we get:

vf = 0.15 kg. 1.03 m/s / 0.13. 0.967 = 1.23 m/s