. At the end of the race a person with a mass of 90 kg. jumps out in front of th
ID: 1975823 • Letter: #
Question
. At the end of the race a person with a mass of 90 kg. jumps out in front of the 230.0kg bobsled which is going 25.0 m/s horizontally. Initially the person is at rest. The person flies up into the air at a 30 degree angle with a speed of 12.0 m/s.a. What is the initial momentum of the person?
b. What is the momentum of the person after they are struck? Give magnitude and direction.
c. What impulse did the bobsled apply to the person?
d. If the collision occurred over a time of 5.00 milliseconds, with what force was the person struck?
e. What is the horizontal component of the person’s final momentum.?
f. Using the value in (e) i.e. the horizontal component, find the final horizontal velocity of the bobsled after the collision. Note: Conservation of momentum implies that it is conserved in both the x and y dimensions. Here the earth gets involved in the y-direction so we don’t deal with it.
Explanation / Answer
momentum conserved in the horizontal plane initial momentum = momentum of car + momentum of man in x axis i.m. = 230*25 - 90*12 cos 30 = 4814.7 a)inotial momemtum = mass *velocity = 90*12 =1081kg m/s final velocity of car +man = 4814.7/320= 15.04m/s b)afetr they struck there is no momentum towards the vertical direction as it dampened so momentum =90*15.04 = 1354.13 in horizontal direction c) as the person di not jump up so vertical impulsse = 90*12sin30 = 540 kg m/s now velocity of man is 15.04 m/s apposite to that previously so change in momentum = (12-(-15.04))*90 = 2433.6 kgm/s d) force = rate of cjange of momentum = 2433.6/.005 = 486720 N e)horizontal = 1354.13 kgm/s f)final horizontal velocity = 15.04 m/s and the car continues its movement.
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