32_new - Ch3-HW2 florin(56950 This print-out should have 16 questions Multiple-choice questions may continue on the next column or page find all

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Ch3-HW2 – florin – (56950) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. And we use the magnitude to help us find the unit vector: ˆr 21 = ~r 21 |~r 21 | 001 (part 1 of 7) 4.0 points In outer space, far from other objects, block 1 of mass 40 kg is at position ~r 1 = h6, 8, 0O m, and block 2 of mass 1000 kg is at position ~r 2 = h19, 12, 0O m. Since the z components of position are zero, the (vector) gravitational force acting on block 2 due to block 1 will be of the form ~ 21 21 21 F gr = hF gr ,x , F gr ,y , 0O N. Find F gr 21 ,x . It helps to make a sketch of the sit- uation. Use G = 6.67 × 10 11 m 3 · kg 1 · s 2 . Correct answer: −1.37839 × 10 8 N. Explanation: The formula for the gravitational force be-tween two objects is ~ Gm 1 m 2 F gr = − |~r| 2 ˆr. We know the masses of the blocks, so we just need to find the magnitude of the distance between them and the unit vector pointing from block 1 to block 2. To find the vector pointing from block 1 to block 2 (let’s call it ~r 21 ), we subtract the position of block 1 from that of block 2: ~r 21 = ~r 2 − ~r 1 = h19, 12, 0O m − h6, 8, 0O m = h13, 4, 0O m . We need the length of this vector and the unit vector pointing in the same direction. The length (or magnitude) is given by ~r 21 = q (13 m) 2 + (4 m) 2 13.6015 m . = h 13, 4, 0O m 13.6015 m = h0.955779, 0.294086, 0O . Now we have everything we need to calcu- ~ late the components of F gr . Let’s go ahead and find all three components using vector algebra. ~ 21 Gm 1 m 2 21 F gr = − |~r 21 | 2 ˆr = ( G)(40 kg)(1000 kg) (13.6015 m) 2 × h0.955779, 0.294086, 0O = h−1.37839 × 10 8 , −4.24119 × 10 9 , 0O N where G = 6.67 × 10 11 m 3 · kg 1 · s 2 . 002 (part 2 of 7) 4.0 points Find F gr 21 ,y . Correct answer: −4.24119 × 10 9 N. Explanation: See the explanation for part 1. 003 (part 3 of 7) 4.0 points At 4.6 s after noon both blocks were at rest at the positions given above. At 4.7 s after noon, what is the (vector) momentum of block 2? Note that the momentum vector will also be of the form ~p 2 f = hp 2 f,x , p 2 f,y , 0O kg · m/s, so start by finding p 2 f,x . Correct answer: −1.37839 × 10 9 N. Explanation: We already know the force on block 2. Us-ing the Momentum Principle, defining block 2 as our system, we can write
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Ch3-HW2 – florin – (56950) 2 2 2 ~ ~p f = ~p i + F net t ~ = F net ( t f t i ) . According to this formula, the x component will be p 2 f,x = 1 . 37839 × 10 8 N (0 . 1 s) = 1 . 37839 × 10 9 kg · m / s .
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This note was uploaded on 04/04/2012 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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32_new - Ch3-HW2 florin(56950 This print-out should have 16 questions Multiple-choice questions may continue on the next column or page find all

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