[ola alharbi]

السلاه عليكم لدي هذا السؤال و أرجو مساعدتي في حله

a golfer hits a golf ball, imparting to it an initial velocity of magnitude 52.2 m/s directed 30 angle
above the horizontal assuming that the mass of the ball is 46.0 g and the club and ball are in contact for 1.20 ms

find the impulse imparted to the ball
find the impulse imparted to the to the club
find the average force exerted on the ball by the club
find the work done on the ball


بجد أحتاج مساعده ضروريه

[ola alharbi]

مافي أي مساعدة ؟

[phet]

تفضل الحل:
السؤال رقم 4 في هذه الصفحة

http://www.askiitians.com/iit-jee-ph...-momentum.aspx

Solution:- The initial momentum pi of the ball will be, pi = mvi To obtain the initial momentum pi of the ball, substitute 46.0 g for the mass of the ball and 52.2 m/s for vi in the equation pi = mvi, pi = mvi = (46.0 g) (52.2 m/s) = (46.0 g×10-3 kg/1 g) (52.2 m/s) = 2.4 kg.m/s = (2.4 kg.m/s) (1 N/1 kg.m/s2) = 2.4 N.s As the final speed of the ball is zero, thus the final momentum pf of the ball will be zero. So, pf = 0 To find out the impulse J imparted to the ball, substitute 2.4 N.s for pi and 0 for pf i n the equation J = pi – pf, J = Δp = pi – pf = 2.4 N.s -0 = 2.4 N.s Therefore the impulse J imparted to the ball would be 2.4 N.s. b) The impulse J imparted to the club is just opposite that imparted to the ball. Therefore the impulse J imparted to the club will be 2.4 N.s. c) To obtain the average force F exerted on the ball by the club, substitute 2.4 N.s for Δp and 1.20 ms for t in the equation F = Δp/ Δt, F = Δp/ Δt = 2.4 N.s/1.20 ms = (2.4 N.s)/((1.20 ms)(10-3 s/1ms)) = 2000 N From the above observation we conclude that, the average force F exerted on the ball by the club would be 2000 N .