ch09-p013

# ch09-p013 - downward angle 34 ° 13(a The net force on the...

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v com = (2.35 i ^ – 1.57j ^ ) t (with SI units understood), since it started at rest. We note that the ratio of the y - component to the x -component (for the velocity vector) does not change with time, and it is that ratio which determines the angle of the velocity vector (by Eq. 3-6), and thus the direction of motion for the center of mass of the system. (c) The last sentence of our answer for part (b) implies that the path of the center-of-mass is a straight line. (d) Eq. 3-6 leads to θ = 34º. The path of the center of mass is therefore straight, at
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Unformatted text preview: downward angle 34 ° . 13. (a) The net force on the system (of total mass m 1 + m 2 ) is m 2 g . Thus, Newton’s second law leads to a = g ( m 2 /( m 1 + m 2 ) ) = 0.4 g . For block1, this acceleration is to the right (the i ^ direction), and for block 2 this is an acceleration downward (the –j ^ direction). Therefore, Eq. 9-18 gives a com → = m 1 a 1 → + m 2 a 2 → m 1 + m 2 = (0.6)(0.4 g i ^ ) + (0.4)(–0.4 g j ^ ) 0.6 + 0.4 = (2.35 i ^ – 1.57 j ^ ) m/s 2 . (b) Integrating Eq. 4-16, we obtain...
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## This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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