Two astronauts on opposite ends of a spaceship are comparing lunches. One has an

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 = 1.10 m/s . The 0.150 kg orange is tossed from astronaut 2 to astronaut 1 with a speed of 1.25 m/s . Unfortunately, the fruits collide, sending the orange off with a speed of 1.05 m/s in the negative y direction. PLEASE show and explain how you got the answer.

Part A What are the final speed and direction of the apple in this case?

Explanation / Answer

(m1) = 0.125 kg, (u1) = ( 1.10 i ) m/s, (m2) = 0.150 kg

(u2) = ( - 1.25 i ) m/s, (v2) = ( - 1.05 j ) m /s

.From conservation of momentum, we have

.m1 u1 + m2 u2 = m1 v1 + m2 v2

.m1 v1 = (  m1 u1 + m2 u2 ) - m2 v2

.   = ( 0.125 * 1.10 i + 0.150 * -1.25 i ) - ( 0.150 * -1.05 j )

= - 0.05 i +0.1575 j

.then the velocity is,

v1 =[ - 0.05 i +0.1575 j] /0.125 = -0.4 i + 1.26 j

magnitude of the velocity is :

      v1 = sqrt [ (-0.4)2   + (1.26)2 ] = 1.32 m/s

Direction is :    = tan-1  ( 1.26 / -0.4 ) = -72.38o = 17.6o

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