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Skater A, with a mass of 72.0 kg, is moving at 11.0 m/s [S] when he collides with skater B, with a mass of 42.0 kg, moving at 14.0 m/s [E]. The collision is completely inelastic and the two skaters move off together after the collision. Find the velocity of the skaters, right after the collision. Can you please explain why in the calculation vAy = - 11.0 m/s.

Skater A, with a mass of 72.0 kg, is moving at 11.0 m/s [S] when he collides with skater B, with a mass of 42.0 kg, moving at 14.0 m/s [E]. The collision is completely inelastic and the two skaters move off together after the collision. Find the velocity of the skaters, right after the collision. Can you please explain why in the calculation vAy = - 11.0 m/s.

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Skater A, with a mass of 72.0 kg, is moving at 11.0 m/s [S] when he collides with skater B, with a mass of 42.0 kg, moving at 14.0 m/s [E]. The collision is completely inelastic and the two skaters move off together after the collision. Find the velocity of the skaters, right after the collision.

Can you please explain why in the calculation vAy = - 11.0 m/s.

Thank you.

Draw the given velocity vectors: V = 0 %D 1х V. ly =-11.0 2x +14.0 Vzy = 0 Now use conservation of momentum for the components. For the x-components: PTox PT m,V 1ox + m,v20x = (m,+ m,)v (72.0)(0) + (42.0)(14.0) = (114)v %3D v. = 5.158 m/s %3D fx For the y-components: PToy Prfy m,v + m,y20 = (m,+ m,)V %3D 2 2oy loy (72)(-11.0) + (42)(0) = (114)v, %3D fy V = -6.947 m/s Find the final velocity of the two skaters, using the following triangle: 1 5.158 m/s (0302=(010 காயறிபல 6.947 m/s V. f v; = V(5.158 m/s) +(6.947 m/s) %3D Vf = 8.65 m/s To find the direction: 6.947 m/s tane = 5.158 m/s 0 = 53.4° %D The final velocity of the two skaters is 8.65 m/s [E 53.4° S]. (6 marks)
Transcribed Image Text:Draw the given velocity vectors: V = 0 %D 1х V. ly =-11.0 2x +14.0 Vzy = 0 Now use conservation of momentum for the components. For the x-components: PTox PT m,V 1ox + m,v20x = (m,+ m,)v (72.0)(0) + (42.0)(14.0) = (114)v %3D v. = 5.158 m/s %3D fx For the y-components: PToy Prfy m,v + m,y20 = (m,+ m,)V %3D 2 2oy loy (72)(-11.0) + (42)(0) = (114)v, %3D fy V = -6.947 m/s Find the final velocity of the two skaters, using the following triangle: 1 5.158 m/s (0302=(010 காயறிபல 6.947 m/s V. f v; = V(5.158 m/s) +(6.947 m/s) %3D Vf = 8.65 m/s To find the direction: 6.947 m/s tane = 5.158 m/s 0 = 53.4° %D The final velocity of the two skaters is 8.65 m/s [E 53.4° S]. (6 marks)
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