F p r the impulse acting on it is 3 ˆ 74 10 n si f i

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f p   r The impulse acting on it is 3 ˆ ( 7.4 10 N s)i. f i J p p   (c) The average force on the car during the turn is ) )( 1600 ( 6 . 4 ) )( / 7400 ( j i N s j i s m kg t J t p F avg and its magnitude is 3 avg 1600N 2 2.3 10 N. F (d) The average force during the collision with the tree is 4 avg 3 ˆ 7400kg m s i ˆ 2.1 10 N i 350 10 s J F t   r r and its magnitude is 4 avg 2.1 10 N F . (e) As shown in (c), the average force during the turn, in unit vector notation, is avg ˆ ˆ 1600N i j F r . Note: During the turn, the average force avg F r is in the same direction as J r , or p r . Its x and y components have equal magnitudes. The x component is positive and the y component is negative, so the force is 45° below the positive x axis. 11. (a) We place the origin of a coordinate system at the center of the pulley, with the x axis horizontal and to the right and with the y axis downward. The center of mass is halfway between the containers, at x = 0 and y = , where is the vertical distance from the pulley center to either
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of the containers. Since the diameter of the pulley is 50 mm, the center of mass is at a horizontal distance of 25 mm from each container. (b) Suppose 20 g is transferred from the container on the left to the container on the right. The container on the left has mass m 1 = 480 g and is at x 1 = 25 mm. The container on the right has mass m 2 = 520 g and is at x 2 = +25 mm. The x coordinate of the center of mass is then . 0 . 1 520 480 ) 25 )( 520 ( ) 25 )( 480 ( 2 1 2 2 1 1 mm g g mm g mm g m m x m x m x com The y coordinate is still . The center of mass is 26 mm from the lighter container, along the line that joins the bodies. (c) When they are released the heavier container moves downward and the lighter container moves upward, so the center of mass, which must remain closer to the heavier container, moves downward. (d) Because the containers are connected by the string, which runs over the pulley, their accelerations have the same magnitude but are in opposite directions. If a is the acceleration of m 2 , then a is the acceleration of m 1 . The acceleration of the center of mass is . ) ( 2 1 1 2 2 1 2 1 m m m m a m m a m a m a com We must resor t to Newton’s second law to find the acceleration of each container. The force of gravity m 1 g , down, and the tension force of the string T , up, act on the lighter container. The second law for it is m 1 g T = m 1 a . The negative sign appears because a is the acceleration of the heavier container. The same forces act on the heavier container and for it the second law is m 2 g T = m 2 a . The first equation gives T = m 1 g + m 1 a . This is substituted into the second equation to obtain m 2 g m 1 g m 1 a = m 2 a , so a = ( m 2 m 1 ) g /( m 1 + m 2 ).
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