Two astronauts on opposite ends of a spaceship are comparing lunches. One has an
ID: 1773684 • Letter: T
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 = 1.12 m/s . The 0.170 kg orange is tossed from astronaut 2 to astronaut 1 with a speed of 1.20 m/s . Unfortunately, the fruits collide, sending the orange off with a speed of 0.986 m/s in the negative y direction. What are the final speed and direction of the apple in this case? counterclockwise from the positive x axis?
Explanation / Answer
Given
m1 = 0.13 kg
u1 = 1.12 m/s i
m2 = 0.17 kg
u2 = - 1.20 m/s i
v2 = - 0.986 m/s j
Applying conservation of kinetic energy
0.5 m1 u1^2 + 0.5 m2 u2^2 = 0.5 m1 v1^2 + 0.5 m2 v2^2
0.13*1.12^2 + 0.17*1.20^2 = 0.13 v1^2 + 0.16*0.986^2
v1 = sqrt((0.13*1.12^2 + 0.17*1.20^2- 0.17*0.986^2)/0.13) = 1.366 m/s
Final speed of apple = 1.366 m/s
Applying conservation of momentum
m1 u1 + m2 u2 = m1 v1 + m2 v2
0.13 * 1.12 i - 0.17*1.20 i = 0.13* v1 - 0.17*0.986 j
v1 = (0.13 * 1.12 i - 0.17*1.20 i + 0.17*0.986 j ) /0.13
v1 = (- 0.0584 i + 0.17*0.986 j ) / 0.13 = - 0.449 i + 1.289 j
Direction of apple = 180 - arctan(1.289/0.449) = 109.2 0 counterclockwise from the positive x axis
theta = 109.2 0 counterclockwise from the positive x axis