Question

Two astronauts on opposite ends of a space ship are comparing lunches. One has an apple, the other has an orange. They decide to trade. Astronaut 1 tosses the 0.160 kg apple toward astronaut 2 with a speed of 2.15 m/s. The 0.130 kg orange is tossed from astronaut 2 to astronaut 1 with a speed of 1.05 m/s. Suppose that after the collision the orange moves exactly in the negative y direction with a speed of 1.52 m/s. What is the final speed and direction of the apple in this case?

Answer 1

Answer:

$v_{1f}$  =  - 0.0519 m/s, and the apple direction is in the negative direction of the x-axis

Explanation:

This problem should be used moment conservation. Let us form a system composed of the apple and the orange, for this system the forces during the crash are internal, therefore, the moment is preserved. Let's write the moment one moment before the crash and another moment after the crash

Initial. Before shock

    p₀ = m₁ v₁₀ + m₂v₂₀

Final. After the crash

    pf = m₁ $v_{1f}$ + m₂ $v_{2f}$

   p₀ = pf

   m₁ v₁₀ + m₂v₂₀ = m₁ $v_{1f}$  + m₂ $v_{2f}$

They give us several data, the mass of the apple m₁ = 0.160 kg, its initial velocity v₁₀ = 2.15 m/s, the mass of the orange m₂ = 0.130 kg, its initial velocity v₂₀ = 1.05 m/s and the final velocity $v_{2f}$  = - 1.52 m/s.

Let us analyze the signs of the velocities for the orange, as it is the lightest body in the crash, the direction of its velocity must be reversed, if the final velocity is negative, this implies that the initial velocity was positive, bone moves to the right of the x-axis. . The only way that the shock is that the apple has an initial speed in the negative direction.

Let's calculate the finan speed of the Massana

    m₁ $v_{1f}$  = m₁ v₁₀ + m₂ v₂₀ - m₂ $v_{2f}$

    0.160 $v_{1f}$  = 0.160 (-2.15) + 0.130 1.05 - 0.130 (-1.52)

    $v_{1f}$ = -0.0099 / 0.160

   $v_{1f}$  =  - 0.0519 m/s

The apple is in the negative direction of the x-axis, but with much lower speed