two african swallows fly toward each other carrying coconuts. the first swallow

Question

two african swallows fly toward each other carrying coconuts. the first swallow is flying north horizontally with a speed of 20 m/s. the second swallow is flying at the same height as the first and in opposite direction with speed of 15m/s. the mass of first swallow is 0.270 kg and the mass of his coconut is 0.80 kg, the second swallows mass is 0.220 kg and her coconuts mass is .70kg. the swallows collide and lose their coconuts. immediately after collision the o.80 kg coconut travel 10 degrees west of south with speed 13m/s and the 0.70 kg coconut moves 30 degrees east of north with a speed of 14 m/s. the two birds are tangled up with each other and stop clapping wings what is the velocity of the birds immediately after the collision

Explanation / Answer

KE balance 0.5*0.92*(15)^2 + 0.5*(1.07)*(20)^2 = (0.5)*(0.8)*(13)^2 + (0.5)*(0.7)*(14)^2 + 0.5(.27)*(v1)^2 + 0.5*(.22)*(v2)^2 103.5 + 214 = 67.6 + 68.6 + .135v1^2 + .11v2^2 181.3= .135v1^2 + .11v2^2 v2= root[(181.3 - 0.135v1^2)/(0.11)] Momentum balance: -0.8*13*cos10 + 0.7*(14)*cos30 + (.27)*(v1) - (.22)*(v2) = -(0.92)*(15) + (1.07)*(20) -10.24 + 8.48 + .27v1 -.22v2 = -13.8 + 2.14 (9.9 + 0.27v1)/0.22 = v2 solve these two equations and get v1 and v2

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