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1 April, 11:02

Jack (mass 55.0 kg) is sliding due east with speed 8.00 m>s on the surface of a frozen pond. he collides with jill (mass 48.0 kg), who is initially at rest. after the collision, jack is traveling at 5.00 m>s in a direction 34.0° north of east. what is jill's velocity (magnitude and direction) after the collision? ignore friction

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  1. 1 April, 11:16
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    Jill will have an angle of 36.0° south of east with a velocity of 5.46 m/s This problem deals with the conservation of momentum. The momentum before and after the collision will remain constant and is a vector quantity. So let's calculate the momentum before the collision. 55.0 kg * 8.00 m/s = 440 kg*m/s So we have a vector with a magnitude of 440 kg*m/s and an angle of 0° before the collision. After the collision, we have Jack's vector being 55.0 kg * 5.00 m/s = 275 kg*m/s So his new vector has a magnitude of 275 kg*m/s and an angle of 34.0° So what we have is a side angle side (SAS configuration of a triangle where one side has a length of 440, the other side has a length of 275 and the included angle is 34.0°. We need to calculate the length of the opposite side. Using the law of cosines, we have c^2 = a^2 + b^2 - 2ab cos (C) Substituting known values gives. c^2 = 440^2 + 275^2 - 2*440*275*cos (34) c^2 = 193600 + 75625 - 242000*0.829037573 c^2 = 269225 - 200627.0926 c^2 = 68597.90744 c = 261.9120223 So the magnitude of Jill's momentum will be 261.912 kg*m/s. And dividing by her mass, gives us a velocity of 261.912 kg*m/s / 48.0 kg = 5.4565 m/s Her angle will be south of due east. The measurement of the angle will be the same as the angle opposite to the side of length 275. And using the law of sines, we get sin (34.0) / 261.9120223 = sin (X) / 275 0.559192903/261.9120223 = sin (X) / 275 0.002135041 = sin (X) / 275 0.587136272 = sin (X) 35.95405166 = X So Jill will have an angle of 36.0° south of east with a velocity of 5.46 m/s
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