an 8kg mass moving east at 15 m/s of a frictionless horizontal surface collides
ID: 2051027 • Letter: A
Question
an 8kg mass moving east at 15 m/s of a frictionless horizontal surface collides with a 10 kg mass that is initially at rest. After the collision, the 8 kg mass moves south at 4 m/s. What is the velocity of the 10 kg mass after the colliosion?Found some help online, but the answer doesn't work out and I'm not sure I'm doing this right anyway. Posted last night, and someone wrote a very nice answer, but it doesn't work ot (and I can't make sense of it, the equations aren't balanced). I really want to understand this.
SN direction
0= 8(5) + 10v
35=10v
-3.5=v north (which is why it is negative)
EW 8(15) = 10 v
120 = 10 v
12 =v east
magnitude 9right triangle)
4^2 + 12 ^2 =160 then sqrt = 12.649
tan -1 = 4/12 = 18.43
it's supposed to be 12.4 at 14.9 degrees north of east
Explanation / Answer
initial momentum of the total system in east direction = 8*15 + 0 = 120 after collision 8kg mass moves south with a velocity of 4 m/s 8kg does not have any component in east direction but initially there was some momentum in east direction so 10 kg mass must balance that momentum 10*ve = 120 ve = 12 10 kg mass has a component of 12 m/s in east direction now in south direction there wasn't any momentum initially so 10 kg must have momentum component in north direction to balance the momentum of 8 kg 10vn = 8*4 vn = 3.2 m/s total velocity after collision = sqrt(12^2 + 3.2^2) = 12.42 at an angle tan-1(3.2/12) = 14.93 above teh east that is 14.93 north of east I hope this is clear
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