Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
A golfer hits a golf ball, imparting to it an initial velocity of magnitude 52.2 m/s directed 30° 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 (a) the impulse imparted to the ball, (b) the impulse imparted to the club, and (c) the average force exerted on the ball by the club.
The initial momentum pi of the ball will be,pi = mviTo 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.sAs the final speed of the ball is zero, thus the final momentum pf of the ball will be zero.So, pf = 0
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -