YF_ISM_08

YF_ISM_08 - MOMENTUM, IMPULSE, AND COLLISIONS 8 8.1....

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8-1 M OMENTUM , I MPULSE , AND C OLLISIONS 8.1. IDENTIFY and SET UP: . pm v = 2 1 2 . Km v = EXECUTE: (a) 5 (10,000 kg)(12.0 m/s) 1.20 10 kg m/s p == × (b) (i) 5 1.20 10 kg m/s 60.0 m/s 2000 kg p v m ×⋅ = . (ii) 22 11 TT S U VS U V mv m v = , so T SUV T SUV 10,000 kg (12.0 m/s) 26.8 m/s 2000 kg m vv m = EVALUATE: The SUV must have less speed to have the same kinetic energy as the truck than to have the same momentum as the truck. 8.2. IDENTIFY: Example 8.1 shows that the two iceboats have the same kinetic energy at the finish line. 2 1 2 v = . v = . SET UP: Let A be the iceboat with mass m and let B be the iceboat with mass 2m , so 2 BA mm = . EXECUTE: AB KK = gives mv mv = . 2 B B A m v m . AA A v = . () (2 ) / 2 2 2 BB B A A A A A v mv m v p = = . EVALUATE: The more massive boat must have less speed but greater momentum than the other boat in order to have the same kinetic energy. 8.3. IDENTIFY and SET UP: v = . 2 1 2 v = . EXECUTE: (a) p v m = and 2 2 1 2 2 pp ⎛⎞ ⎜⎟ ⎝⎠ . (b) cb = and the result from part (a) gives = . b bc c c c 0.145 kg 1.90 0.040 kg m p p m = . The baseball has the greater magnitude of momentum. / 0.526 = . (c) 2 2 K = so mw = gives ww mK = . wm g = , so wK = . m m m w 700 N 1.56 450 N w K K w = . The woman has greater kinetic energy. / 0.641 = . EVALUATE: For equal kinetic energy, the more massive object has the greater momentum. For equal momenta, the less massive object has the greater kinetic energy. 8.4. IDENTIFY: Each momentum component is the mass times the corresponding velocity component. SET UP: Let + x be along the horizontal motion of the shotput. Let + y be vertically upward. cos x θ = , sin y = . EXECUTE: The horizontal component of the initial momentum is cos (7.30 kg)(15.0 m/s)cos40.0 83.9 kg m/s xx vm v = = ° . The vertical component of the initial momentum is sin (7.30 kg)(15.0 m/s)sin40.0 70.4 kg m/s yy v = = ° EVALUATE: The initial momentum is directed at 40.0 ° above the horizontal. 8
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8-2 Chapter 8 8.5. IDENTIFY: For each object, m GG p =v and 2 1 2 Km v = . The total momentum is the vector sum of the momenta of each object. The total kinetic energy is the scalar sum of the kinetic energies of each object. SET UP: Let object A be the 110 kg lineman and object B the 125 kg lineman. Let + x be the object to the right, so 2.75 m/s Ax v =+ and 2.60 m/s Bx v =− . EXECUTE: (a) (110 kg)(2.75 m/s) (125 kg)( 2.60 m/s) 22.5 kg m/s xA A xB B x Pm v m v =+= + = . The net momentum has magnitude 22.5 kg m/s and is directed to the left. (b) 22 2 2 111 1 222 2 (110 kg)(2.75 m/s) (125 kg)(2.60 m/s) 838 J AA BB v m v = + EVALUATE: The kinetic energy of an object is a scalar and is never negative. It depends only on the magnitude of the velocity of the object, not on its direction. The momentum of an object is a vector and has both magnitude and direction. When two objects are in motion, their total kinetic energy is greater than the kinetic energy of either one. But if they are moving in opposite directions, the net momentum of the system has a smaller magnitude than the magnitude of the momentum of either object.
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YF_ISM_08 - MOMENTUM, IMPULSE, AND COLLISIONS 8 8.1....

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