Question

A 0.060-kg tennis ball, initially moving at a speed of 12 m/s, is struck by a racket. The direction of the ball is reversed and its speed is increased to 18 m/s. A high-speed video of the impact shows that the duration of the contact between the ball and the racket was 0.050 s. What is the magnitude of the average force that was exerted on the ball by the racket while they were in contact? (MAGNITUDE)

A 36 N B 32 N C 7.2 N D 1.4 N E 0.72 N

Answer 1

Mass of tennis ball = M = 0.060 Kg Let ball is initially moving towards right and after impact it changes its direction to left. Initial speed of ball = Vi = 12 m/s final speed of ball = V₂ = 18 m/s (Take right direction as positive and left direction as negative) Duration of contact between ball and rocket = At=0.050s Hence, Initial velocity of ball = 0 = +12 m/s Final velocity of ball = V2 = - 18 mis (Final (Initial) do Momentum = mass x reloc'ty = Initial momentum of ball = t = mv final momentum of ball = 22=mV2 Change in momentum of ball = Ap=/Final momentum B - Initial momentum) ab = 2 - 8 - Ap=mV2 - mi 3) sto = m ( 12-T ) -) s6 = 0.060 Kg (-18 m/s-12 m)s) Ap = 0.060 Kg (-30 m/s) =) sſ = -1.8. kg ms! Now, Average force is given by = Change in momentum Duration of contact Farg Du - Farg = A - Average force exerted by ball on rocket is

-l. - 2 = - 3 S. Farg ball = -1.8 kg m5' - Farg ball = -36 kg ms? 0.osos =) Farg ball =-36 N. [oo lkg m5² = 1N] Now, by newtons third law, we have - Force exerted by rocket on ball = -(force exerted by 1 ball on rocket) =) Farg rocket = - Farg ball = Fargrocket =-(-36N) =) Farg rocket = 36N . So, magnitude of the arerage force exerted on ball by rocket = | Farg rocket) = Farg rocket = 36 N . Hence, answer is (A) 36N