Question

Two skydivers are holding on to each other while falling straight down at a common terminal speed of 63.10 m/s. Suddenly, they push away from each other. Immediately after separation, the first skydiver (who has a mass of 89.30 kg) has the following velocity components (with "straight down" corresponding to the positive z-axis): v_1, x = 4.930 m/s v_1, y = 4.750 m/s v_1, z = 63.10 m/s What are the x- and y-components of the velocity of the second skydiver, whose mass is 57.70 kg, immediately after separation? What is the change in kinetic energy of the system?

Answer 1

For first question answer as following

We have v_1x = 4.93m/s; v_1y = 4.75m/s ; v_1z =63.1m/s

            First sky driver mass, m₁ = 89.30kg

            Second sky driver mass, m₂ =57.7kg

Now

v_2x= -v_1x x (m₁/m₂) = - 4.93 x (89.3 /57.7) = -4.93 x1.548 =-7.63 m/s

v_2y = -v_1y x (m₁/m₂) = - 4.75 x (89.3 /57.7) = -4.75 x 1.548 =-7.353m/s

v_2z = ((m₁+m₂)63.1-v_1z m₁)/ m₂ = [(147x63.1) – (63.1 x89.30)]/57.7

            =[9275.7 -5634.83] /57.7 =3640.87/57.7 = 63.10m/s

For second question answer as following

               Initial K.E = ½ m v² = ½ (m₁+m₂) v² = ½ (89.3+57.7) (63.1)²

                                    = 0.5 x147x 3981.61 = 292648.34J

Final K.E = ½ (m₁v₁² +m₂v₂² )

v₁² = v_1x² + v_1y²+v_1z² = (4.93)² +(4.75)² + (63.1)² = 24.31+22.56 +3981.61 =4028.48 m²/s²

v₂² = v_2x²+v_2y²+v_2z² = (-7.63)² + (-7.353)² + (63.1)² =58.22 +54.067+3981.61 =4093.897 m²/s²

final K.E = ½ ( 89.3 x 4028.48 + 57.7 x4093.897) = ½ (359743.264 +236217.86)

= 297980.56 J

ΔK.E = 297980.56- 292648.34J =5332.2J