Chapter 11 SolutionsProblem 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=!Fdt=(650 N)(0.08 s)=52 N"s, in the direction of the force. Problem 6. A proton moving in the positive xdirection at 4.3 Mm/scollides with a nucleus. The collision lasts 0.12 fs, and the average impulsive force is 42î+17ˆ j μN.(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=mvf"mvi=Fav!t,so vf=vi+Fav!t=m=(4.3 Mm/s)î+ (42î+17ˆ j )(10"6N)#(0.12!10"15s)=(1.67!10"27kg)=(7.32î+1.22ˆ j )Mm/s.(b) !f=tan"1(1.22=7.32)=9.48°,CCW from the x-axis, which is the deflection. (Note: vf=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"4N#s,we find I'(55.5 boxes)!(1.25!10"4N#s/box)=6.94!10"3N#s.(The direction of Iis in the direction of F, assumed to be constant.) (b) Fav=I=!t=6.94"10#3N$s=3"10#3s=2.31 N. FIGURE11-18 Problem 11 Solution.
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