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 0.125 kg apple toward astronaut 2 with a speed of vi,1 = 1.09 m/s . The 0.160 kg orange is tossed from astronaut 2 to astronaut 1 with a speed of 1.19 m/s . Unfortunately, the fruits collide, sending the orange off with a speed of 0.982 m/s in the negative y direction.What are the final speed and direction of the apple in this case?

Answer 1

Q1.

Ans- Before collision

           m1= 0.125 Kg   V_i,1 = 1.09x m/s

           m2 = 0.160 Kg   V_i,2 = -1.19x m/s

After collision

            V_f,1 = ?

            V_f,2 = -0.982 y m/s

from conservation of linear momentum along x axis

           m1 V_i,1x + m2 V_i,2x = m1 V_f,1x + m2 V_f,2x

         0.125 * 1.09 + (0.160* (-1.19))= (0.125*V_f,1x) + (0.160*(0))

             V_f,1x = -0.433 m/s

  

from conservation of linear momentum along y axis

           m1 V_i,1y + m2 V_i,2y = m1 V_f,1y + m2 V_f,2y

         0.125 * 0 + (0.160* 0)= (0.125*V_f,1y) + (0.160*(-0.982))

             V_f,1y = 1.257 m/s

So final speed of apple is -

        V_f,1x = sqrroot (V_f,1x ^2   +   V_f,1y ^2) = sqrroot ( ( -0.433)^2   +   (1.257)^2) = 1.33 m/s

Direction of apple -

       

      with x axis (towards astronaut 2 from 1)

So, final speed of apple is 1.33 m/s and Direction of apple is 109.0^o with x axis (towards astronaut 2 from 1).