Two ice skaters, each with a mass of 72.0 kg, are skating at 5.15 m/s when they

Question

Two ice skaters, each with a mass of 72.0 kg, are skating at 5.15 m/s when they collide and stick together. If the angle between their initial directions was 117 degree, determine the components of their combined velocity after the collision. (Let the initial motion of skater 1 be in the positive x direction and the initial motion of skater 2 be at an angle of 117 degree measured counterclockwise from the positive x-axis.) v_x, f = ____________ m/s v_y, f = ____________ m/s A 0.32-kg puck at rest on a horizontal frictionless surface is struck by a 0.22-kg puck moving in the positive x-direction with a speed of 2.2 m/s. After the collision, the 0.22-kg puck has a speed of 1.5 m/s at an angle of theta = 60 degree counterclockwise from the positive x-axis. Determine the velocity of the 0.32-kg puck after the collision. ____________ magnitude After writing a statement of conservation of momentum in the x and y ¡directions, you will have two equations and two unknowns (the magnitude and direction of velocity of the 0.32-kg puck). You can solve the two equations simultaneously in order to obtain the magnitude and direction of velocity of the 0.32-kg puck after the collision, m/s direction _____________ degree (clockwise from the +x-axis) Find the percent of kinetic energy lost in the collision. 100 |Delta KE/KE_i| = _____________ Knowing the mass of each puck and speed of each puck both before and after the collision, we can determine the kinetic energy of each puck before and after the collision and then the percent of the initial kinetic energy lost.%

Explanation / Answer

mass m = 72 kg

Initial velocities u = 5.15 i

                      U = 5.15 cos 117 i + 5.15 sin 117 j

                         = -2.338 i +4.588 j

Where i , j are the unit vectors along X and Y axis respectively.

From law of conservation of linear momentum ,

mu + mU = (m+ m) v

   u + U = 2v

5.15 i -2.338 i +4.588 j = 2v

2.812 i + 4.588 j = 2v

From this final velocity of the skaters after collision v = 1.406 i + 2.294 j

Therefore v xf = 1.406 m/s

              v yf = 2.294 m/s

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