HW10Sol - m2 = 125g 7.50x104kg v1iI= 30.0cm/s nitially Initially 4 Finally v1f6.50x10 kg = 17.5cm/s 35.0m/s m1 = 175g 0.150m/s 25.0m/s m2 = 125g

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Initially Finally 35.0m/s 25.0m/s θ = 55.0 o 7.50x10 4 kg 6.50x10 4 kg 0.150m/s 0.0223m/s v 2i = ? m 1 m 1 m 2 m 2 m 2 = 125g v 1i = 30.0cm/s Initially v 1f = 17.5cm/s m 1 = 175g m 2 = 125g Finally v 2f = ? v 2i = 0 Jim Guinn’s PHYS2211 Assignment #10 Solutions Due Monday, Mar. 1, 2010 1. What is the magnitude and direction of the impulse that would have to be applied to a 65.0g-ball that is moving initially with a velocity of 35.0m/s in the -y direction to change its velocity to 25.0m/s at an angle of 55.0 o above the +x-axis. (Hint: keep in mind that impulse is a vector and solve for the components separately, first.) We know that the impulse is given by J = p = p f - p i , or in terms of the components, J x = p x = p x,f - p x,i , and J y = p y = p y,f - p y,i , we have J x = m v x,f – m v x,i = (0.0650kg) (25.0m/s) cos55.0 o – (0.0650kg) (0) = 0.932kg m/s, and J y = m v y,f – m v y,i = (0.0650kg) (25.0m/s) sin55.0 o – (0.0650kg) (-35.0m/s) = 3.61kg m/s . So
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This note was uploaded on 09/25/2011 for the course PHYSICS 22 taught by Professor Lomant during the Spring '11 term at Georgia Perimeter.

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HW10Sol - m2 = 125g 7.50x104kg v1iI= 30.0cm/s nitially Initially 4 Finally v1f6.50x10 kg = 17.5cm/s 35.0m/s m1 = 175g 0.150m/s 25.0m/s m2 = 125g

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