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S08P2AChap11Sol

# S08P2AChap11Sol - Chapter 11 Solutions Problem 1 What is...

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Chapter 11 Solutions Problem 1. What is the impulse associated with a 650-N force acting for 80 ms? Solution The impulse for a constant force (from Equation 11-1) is I = ! F dt = (650 N)(0.08 s) = 52 N " s , in the direction of the force. Problem 6. A proton moving in the positive x direction at 4.3 Mm/s collides with a nucleus. The collision lasts 0.12 fs, and the average impulsive force is 42 î + 17 ˆ j μ N . (a) Find the velocity of the proton after the collision. (b) Through what angle has the proton’s motion been deflected? Solution (a) Equation 11-2 gives ! p = m v f " m v i = F av ! t , so v f = v i + F av ! t = m = ( 4 . 3 Mm / s ) î + ( 42 î + 17 ˆ j )( 10 " 6 N ) # ( 0 . 12 ! 10 " 15 s ) = ( 1 . 67 ! 10 " 27 kg ) = ( 7 . 32 î + 1 . 22 ˆ j ) Mm / s. (b) ! f = tan " 1 (1.22 = 7.32) = 9.48 ° , CCW from the x -axis, which is the deflection. (Note: v f = 7.42 Mm/s . Problem 11. (a) Estimate the impulse imparted by the force shown in Fig. 11-18. (b) What is the average impulsive force? Solution (a) The impulse is the area under the curve in Fig. 11-18. By counting boxes, each of which has “area” (0.25 ms)(0.5 N) = 1.25 ! 10 " 4 N # s, we find I ' ( 55 . 5 boxes ) ! ( 1 . 25 ! 10 " 4 N # s / box) = 6 . 94 ! 10 " 3 N # s . (The direction of I is in the direction of F , assumed to be constant.) (b) F av = I = ! t = 6 . 94 " 10 # 3 N \$ s = 3 " 10 # 3 s = 2 . 31 N . FIGURE 11-18 Problem 11 Solution.

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