b. after the collision, car B moves to the right at a speed of 4.8 m/s. calculat
ID: 1977839 • Letter: B
Question
b. after the collision, car B moves to the right at a speed of 4.8 m/s.
calculate the speed of car A after the colision.
indicate the direction of motion of car A after the collision (to the left,to the right, none car A is at rest)
is this an elastic collision? yes or no
justify your answer
several students are riding in bumper cars at an amusement park. the combined mass of car A and its occupants is 250.0 kg. the combined mass of car B and its occupants is 200.0 kg. Car A is 15.0 m from car B and moving to the right at 2.0 m/s. when the driver decides to bump into car B which is at rest. a. car A accelerates at 1.5 m/s^2 to a speed of 5 m/s and then continues at constant velocity until it strikes car B. Calculate the total time for car A to travel 15 m. b. after the collision, car B moves to the right at a speed of 4.8 m/s. calculate the speed of car A after the colision. indicate the direction of motion of car A after the collision (to the left,to the right, none car A is at rest) is this an elastic collision? yes or no justify your answerExplanation / Answer
A) first, how long does it take for car A to reach 5m/s v(f) = v(i) + at 5m/s = 2m/s + 1.5m/s^2(t) t = (3/1.5)s = 2s now calculate the distance covered in this time period. d = (1/2)at^2 + v(o)t + d(o) let d(o) = 0 d = (1/2)(1.5)(2)^2 + (2)(2) d = 7m So after 2 sec it has moved 7 meters, and it still has to go 8 more meters before reaching car B. Calculate how long it will take the car to travel 8m at 5m/s. d = v(avg)t d = 8m and v(avg) = 5m/s because it is moving at 5m/s on this interval. so t = 8/5 = 1.6s So the total time = 2 + 1.6 = 3.6sec B) use conservation of momentum. total initial moment = total final momentum MV(i) +mv(i) = MV(f) + mv(f) let m = mass car B and v(i) equal car B's initial velocity let M = mass car A and V(i) equal car A's initial velocity the initial and final velocities are the velocities just before and after the collision. Car A is moving at 5m/s and Car B is motionless just before the collision so: 250(5) + 200(0) = 250(v) + 200(4.8) v = (250(5) - 200(4.8))/250 v = 1.16m/s because the answer for v is positive, this means that the velocity of car A is in the positive direction after the collision. The positive direction is the direction that car A is moving initially (only because I set it up that way, aka i said car A's initial velocity was 5m/s not -5) Because car A is moving 5m/s to the left (towards car B), it is moving to the left.
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