Question
in a game of billiard , a collision occurs between two marble balls of different mass. one ball 400g is initially at rest. After the collision, this ball acquires a velocity of 1.40m/s at an angle of 37 degrees from the original direction of the motion of the other ball (500g), which has a speed of 1.86m/s after the collision. what is the initial speed of the moving ball?
Answers
M1*V1 + M2*V2 = M1*V3 + M2*V4
0.4*0 + 0.50*V2=0.40*1.4[37o] + 0.5*1.86
0 + 0.50V2 = 0.447 + 0.337i + 0.93
0.5V2 = 1.377 + 0.337i
V2 = 2.754 + 0.674i
V2 = sqrt(2.754^2+0.674^2) m/s
0.4*0 + 0.50*V2=0.40*1.4[37o] + 0.5*1.86
0 + 0.50V2 = 0.447 + 0.337i + 0.93
0.5V2 = 1.377 + 0.337i
V2 = 2.754 + 0.674i
V2 = sqrt(2.754^2+0.674^2) m/s