2) what\'s the magnitude of the initial moment of the TRUK? 3) what\'s the angle
ID: 1424590 • Letter: 2
Question
2) what's the magnitude of the initial moment of the TRUK? 3) what's the angle that the car-truck combination travel after the collision? ( give the answer as an angel south of East) 4) What's the magnitude of the momentum of the car-truck combination immediately after the collision? 5)what's the speed of the car-truck combination the collision?A blue car with mass me 405 kg is moving east with a speed of Vc -22 m/s and collides with a purple - 1239 kg that is moving south with a speed of vt - 10 m/s. The two collide and lock together after the collision. h a speed of ve 10 m/s. The two collide and lock together after the collsion 1) What is the magnitude of the initial momentum of the car? kg-m/s Submit) 2) What is the magnitude of the initial momentum of the truck kg-m/s Submit
Explanation / Answer
by the conservation of momentum
momentum before collision = momentum after collision
we use vector representation
momentum before collision = mc*vc i + mt*vt (-j) { i represent east and j represent north so on}
momentum before collision = 405*22 i - 1239*10 j = 8910 i - 12390 j
and momentum after collision = (mc + mt)*vf = 8910 i - 12390 j
final speed of car truck combination (vf) = (8910 i - 12390 j)/(405 + 1239) = 5.42 i - 7.53 j
a)
the magnitude of the initial moment of the car = mc*vc = 405*22 = 8910 kg,m/s
b)
the magnitude of the initial moment of the TRUK = mt*vt = 1239*10 = 12390 kg.m/s
c)
theta = arctan (- 7.53 / 5.42) = -54.25 degree
so the angle that the car-truck combination travel after the collision = 54.25 degree south of east
d)
the magnitude of the momentum of the car-truck combination immediately after the collision = sqrt(8910^2 + 12390^2) = 15261.07 kg.m/s
e)
vf = 5.42 i - 7.53 j
magnitude of vf = sqrt(5.42^2 + 7.53^2) = 9.27 m/s
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