Not having your diagram, I do not know if the first mass is at 50 degrees from the y axis in quadrant 1 or quadrant 2
Therefore I can only tell you what to do
X momentum before = .15 * .9 cos40 (or -cos 40) + .26 *.540
=
X momentum after = .15*.64 cos 11 +or -.26*V2 cos phi
Y momentum before = .15*.9 cos 50 + 0
=
Y momentum after = .15*.9 sin 11 - .26*V2 sin phi
and
(1/2).15 (.9)^2 + (1/2).26(.54)^2
=
(1/2).15(.64)^2 + (1/2).26(V2)^2
Two masses are involved in an elastic collision. The first has a mass of 0.150 kg and an initial velocity of 0.900 m/s directed at an angle of 50 degrees with the positive y-axis as shown below. The second mass, of 0.260 kg moves along the x-axis with an initial velocity of 0.540 m/s. After the collision, the first mass has a velocity of 0.640 m/s at an angle of θ=11 degrees above the positive x-axis. The second mass has an angle φ, directed below the positive x-axis. Find the magnitude and direction of the second mass's final velocity. (answers given: φ = -35.09 degrees, and V2f= 0.703 m/s).
2 answers
its in quadrant 2, going in negative y direction. i already solved the answer though thanks!
1 answer