241_PartUniversity Physics Solution

241_PartUniversity Physics Solution - 7-30 Chapter 7 (e) x1...

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7-30 Chapter 7 (e) 10 3 x x = , and 0 2 0 2 (3 ) 9 Ux x α =− . 2 2 00 1 2 22 0 2 2 () ( ( ) ) 29 9 xx vx Ux Ux mx mm x x x x αα ⎡⎤ ⎛⎞ ⎛⎞⎛⎞ = = ⎢⎥ ⎜⎟ ⎜⎟⎜⎟ ⎝⎠⎝⎠ ⎝⎠ ⎣⎦ . The particle is confined to the region where 1 ( ) < . The maximum speed still occurs at 0 2 x x = , but now the particle will oscillate between 1 x and some minimum value (see part (f)). (f) Note that 1 can be written as 2 0 0 21 2 , 93 3 x x x x ⎤⎡ ⎛⎞⎛⎞⎛⎞ −+ = ⎜⎟⎜⎟⎜⎟ ⎥⎢ ⎝⎠⎝⎠⎝⎠ ⎦⎣ which is zero (and hence the kinetic energy is zero) at 01 3 x = = and 3 0 2 x x = . Thus, when the particle is released from 0 x , it goes on to infinity, and doesn&t reach any maximum distance. When released from 1 x , it oscillates between 3 0 2 x and 0 3 x . EVALUATE: In each case the proton is released from rest and ( ) i E Ux = , where i x is the point where it is released. When 0 i x x = the total energy is zero. When 1 i x x = the total energy is negative. ( ) 0 as x →∞ , so for this case the proton can’t reach x and the maximum x it can have is limited. Figure 7.87
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8-1 M OMENTUM , I MPULSE , AND C OLLISIONS 8.1. IDENTIFY and SET UP: . p mv = 2 1 2 . K mv = 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 K mv = . p mv = . SET UP: Let A be the iceboat with mass m and let B be the iceboat with mass 2m , so 2 B A mm = . EXECUTE: AB K K = gives mv mv = . 2 B B A m v m . AA A p = . ( ) (2 ) / 2 2 2 B BB A A A p 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: p mv = . 2 1 2 K mv = . EXECUTE: (a) p v m = and 2 2 1 2 2 p p Km ⎛⎞ ⎜⎟ ⎝⎠ . (b) cb K K = and the result from part (a) gives p p = . b b cc c c 0.145 kg 1.90 0.040 kg m p pp p m = . The baseball has the greater magnitude of momentum. / 0.526 = . (c) 2 2 pm K = so mw p p = gives ww mK = . wm g = , so wK = . m m m w 700 N 1.56 450 N w K KK 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.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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241_PartUniversity Physics Solution - 7-30 Chapter 7 (e) x1...

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