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mid term 1 practice 2

# mid term 1 practice 2 - brown(twb493 extra credit 02...

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brown (twb493) – extra credit 02 – turner – (57340) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Two rocks are tied together with a string of negligible mass, and thrown into the air. At a particular instant, rock 1, which has a mass of 0 . 1 kg, is moving with velocity v 1 = ( 7 , 5 , 4 ) m/s, and rock 2, which has a mass of 0 . 25 kg, is moving with velocity v 2 = (− 7 , 4 , 4 ) m/s. What is the magnitude of the momentum of the system of the two rocks? Correct answer: 1 . 82003 kg · m / s. Explanation: Let : m 1 = 0 . 1 kg , m 2 = 0 . 25 kg , v 1 x = 7 m / s , v 1 y = 5 m / s , v 1 z = 4 m / s , v 2 x = 7 m / s , v 2 y = 4 m / s , and v 2 z = 4 m / s . The total momentum of the system of the two rocks is vectorp total = vectorp 1 + vectorp 2 , so The magnitude of the total momentum is | vectorp total | = | vectorp 1 + vectorp 2 | = { ( p 1 x + p 2 x ) 2 + ( p 1 y + p 2 y ) 2 +( p 1 z + p 2 z ) 2 } 1 / 2 = { ( m 1 v 1 x + m 2 v 2 x ) 2 + ( m 1 v 1 y + m 2 v 2 y ) 2 +( m 1 v 1 z + m 2 v 2 z ) 2 } 1 / 2 = braceleftBig [(0 . 1 kg)(7 m / s) +(0 . 25 kg)( 7 m / s)] 2 + [(0 . 1 kg)( 5 m / s) +(0 . 25 kg)(4 m / s)] 2 + [(0 . 1 kg)(4 m / s) +(0 . 25 kg)(4 m / s)] 2 bracerightBig 1 / 2 = 1 . 82003 kg · m / s . 002 10.0points A charged particle is acted upon by a force of magnitude 120N over a time t. Given that the initial momentum is (15 , 10 , 0) kg.m/s and that the final momen- tum is (20 , 20 , 10) kg.m/s , find t and the force F (in Newtons). 1. 0.125s, (80 , 40 , 40) N 2. 0.25s, (20 , 40 , 40) N 3. 0.125s, ( 80 , 40 , 80) N 4. 0.125s, ( 40 , 80 , 80) N 5. 0.25s, (20 , 40 , 40) N 6. 0.125s, (40 , 80 , 80) N correct 7. 0.125s, (40 , 80 , 80) N Explanation: Given that vector p i = (15 , 10 , 0) kg.m/s vector p f = (20 , 20 , 10) kg.m/s vector | F | = 120 N We use the formula for the magnitude of the impulse = | vector p f vector p i | = | vector F | t t = ( | vector p f vector p i | ) | vector F | = 0 . 125 s vector F = ( vector p f vector p i ) t = (40 , 80 , 80) N 003 10.0points

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brown (twb493) – extra credit 02 – turner – (57340) 2 Consider four vectors vector F 1 , vector F 2 , vector F 3 , and vector F 4 , with magnitudes are F 1 = 47 N, F 2 = 22 N, F 3 = 18 N, and F 4 = 46 N, and θ 1 = 120 , θ 2 = 150 , θ 3 = 24 , and θ 4 = 72 , measured from the positive x axis with the counter- clockwise angular direction as positive.
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mid term 1 practice 2 - brown(twb493 extra credit 02...

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