256_PartUniversity Physics Solution

256_PartUniversity Physics Solution - Momentum, Impulse,...

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Momentum, Impulse, and Collisions 8-15 (b) 4 11 (1200 kg)(12.0 m/s) (1800 kg)(20.0 m/s) 5.04 10 kg m/s. xA A B B Pm v m v =+= + = × (c) ,, cm, (1200 kg)(12.0 m/s) (1800 kg)(20.0 m/s) 16.8 m/s. 1200 kg 1800 kg AA x BBx x AB mv v mm + + == = ++ (d) 4 cm- (3000 kg)(16.8 m/s) 5.04 10 kg m/s xx PM v = ×⋅ , the same as in part (b). EVALUATE: The total momentum can be calculated either as the vector sum of the momenta of the individual objects in the system, or as the total mass of the system times the velocity of the center of mass. 8.51. IDENTIFY: Use Eq. 8.28 to find the x and y coordinates of the center of mass of the machine part for each configuration of the part. In calculating the center of mass of the machine part, each uniform bar can be represented by a point mass at its geometrical center. SET UP: Use coordinates with the axis at the hinge and the + x and + y axes along the horizontal and vertical bars in the figure in the problem. Let ii (,) x y and ff (, ) x y be the coordinates of the bar before and after the vertical bar is pivoted. Let object 1 be the horizontal bar, object 2 be the vertical bar and 3 be the ball. EXECUTE: 2 2 33 i 123 (4.00 kg)(0.750 m) 0 0 0.333 m 4.00 kg 3.00 kg 2.00 kg mx x mmm + + = + + . i 0 (3.00 kg)(0.900 m) (2.00 kg)(1.80 m) 0.700 m 9.00 kg my my my y + + = . f (4.00 kg)(0.750 m) (3.00 kg)( 0.900 m) (2.00 kg)( 1.80 m) 0.366 m 9.00 kg x +− . f 0 y = . fi 0.700 m −= and 0.700 m yy . The center of mass moves 0.700 m to the right and 0.700 m upward. EVALUATE: The vertical bar moves upward and to the right so it is sensible for the center of mass of the machine part to move in these directions. 8.52. (a) IDENTIFY: Use Eq. 8.28. SET UP: The target variable is 1 . m EXECUTE: cm 2.0 m, x = cm 0 y = ( ) ( )( ) () 1 cm 12 1 1 0 0.10 kg 8.0 m 0.80 kg m 0.10 kg 0.10 kg m x m m + + = + . cm 2.0 m x = gives 1 0.80 kg m 2.0 m 0.10 kg m = + . 1 0.80 kg m 0.10 kg 0.40 kg. 2.0 m m += = 1 0.30 kg. m = EVALUATE: The cm is closer to 1 m so its mass is larger then 2 . m (b) IDENTIFY: Use Eq. 8.32 to calculate . P ! SET UP: cm & 5.0 m/s . v j ! 5 ( ) ( ) cm && 0.10 kg 0.30 kg 5.0 m/s 2.0 kg m/s . M + Pv i i ! ! 55 5 (c) IDENTIFY: Use Eq. 8.31. SET UP: cm . + + vv v !! ! 5 The target variable is 1 . v ! Particle 2 at rest says 2 0. v = EXECUTE: ( ) 1c m 1 0.30 kg 0.10 kg 5.00 m/s 6.7 m/s . 0.30 kg m ⎛⎞ ⎜⎟ ⎝⎠ i i 5 EVALUATE: Using the result of part (c) we can calculate 1 p ! and 2 p ! and show that P ! as calculated in part (b) does equal . pp 1 8.53. IDENTIFY: There is no net external force on the system of James, Ramon and the rope and the momentum of the system is conserved and the velocity of its center of mass is constant. Initially there is no motion, and the velocity of the center of mass remains zero after Ramon has started to move.
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8-16 Chapter 8 SET UP: Let + x be in the direction of Ramon&s motion. Ramon has mass R 60.0 kg m = and James has mass J 90.0 kg m = . EXECUTE: RR JJ cm- RJ 0 xx x mv v mm + == + . R JR J 60.0 kg (0.700 m/s) 0.47 m/s 90.0 kg m vv m ⎛⎞ =− ⎜⎟ ⎝⎠ . James& speed is 0.47 m/s. EVALUATE: As they move, the two men have momenta that are equal in magnitude and opposite in direction, and the total momentum of the system is zero. Also, Example 8.14 shows that Ramon moves farther than James in the same time interval. This is consistent with Ramon having a greater speed.
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256_PartUniversity Physics Solution - Momentum, Impulse,...

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