1. The east (e) and south (s) momentum vectors remain the same, and the velocity ratio of those components is the same as the momentum ratio.
(MVs)/MVe) = Vs/Ve = (1x3)/(2x2.5) = 0.6
The final angle, measured south of east is arctangent 0.6 = 31.0 degrees
The final speed is
|V| = (Final momentum)/m
sqrt(3^2 + 5^2)/3 = 1.94 m/s
2. Let Ft be the string tension force. I will assume that the force Fh = 13.8 N that holds it at that angle is horizontal ("to the right"). g, h and t are subscripts, not variables.
The FBD tells you that
Ft sin 24 = Fh = 13.8
Ft cos 24 = Fg
Take the ratio to eliminate Ft.
tan 24 = 13.8/Fg
Fg = 31.0 Newtons
Since Fg is the weight (M*g), divide by g = 9.8 m/s^2 to get the mass M.
End of the semester, working on the final review...
I have a couple of questions I am stuck on right now. My notes on these questions seem to have vanished into the void...
1. "A 2.0kg puck travelling due east at 2.5m/s collides with a 1.0kg puck travelling due south at 3.0m/s. They stick together on impact. What is the resultant direction of the combined pucks?"
Now, I added up the vectors and know that they will be travelling S of E, but don't know how to find the degrees?
2. "A mass suspended by a string is held 24 degrees from vertical by a forse of 13.8N. Find the mass."
My FBD has a mass with an Ft, a force of 13.8N pulling to the right, and an Fg heading straight down. My Fg=mg, but now I am sort of stuck.
Please help!:)
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