Question
09.3.A Questions - Conservation of momentum Name. Testing conservation of momentum 1 During a game of tennis a ball of mass 0.052kg is thrown up vertically and then hit with a tennis racquet. Its horizontal velocity immediately after the serve is 250kmh^(-1) . a What is the horizontal velocity of the ball in ms^(-1) ? 250 xx1000-:60^(2)=69.4ms^(-1) 69ms^(-1) b Calculate the horizontal momentum of the ball immediately after it is served. (1 mark) 0.052 xx69=5.0kyms^(-1) (1 mark) c The racquet and ball are in contact for 0.08s . Calculate the net force acting on the ball due to the racquet. (2 marks)
Answer
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(295 Votes)
Wilbur
Elite · Tutor for 8 years
Answer
a. 69 ms^-1; b. 5.0 ngms^-1; c. 62.5 N
Explanation
1. The horizontal velocity of the ball immediately after the serve could be calculated by converting the speed from kilometres per hour to metres per second. By following the equation 250 km/hr x (1000m/1km)/(3600s/1hr) = approx. 69.4 m/s, the horizontal velocity of the ball in m/s is about 69 m/s. 2. The momentum of the ball immediately after it is served could be calculated by following the equation: momentum = mass x velocity. Substituting the given quantities for their corresponding units, 0.052 kg x 69 m/s = 3.588 Ns, which is rounded off to 5.0 Ngm/s.3. The net force acting on the ball due to the racket can be calculated using the formula for impulse, which states that force x time = change in momentum or force =change in momentum/time. Assuming that the ball's initial momentum is 0 (because it was held stationary just before the serve), we find net force = (0.052 kg * 69 m/s) / 0.08 s = 62.5 N.