One billiard ball is shot east at 1.66 m/s. A second, identical billiard ball is

Question

One billiard ball is shot east at 1.66 m/s. A second, identical billiard ball is shot west at 0.81 m/s. The balls have a glancing collision, not a head-on collision, deflecting the second ball by 90° and sending it north at 1.44 m/s. What is the speed of the first ball after the collision? What is the direction of the first ball after the collision? Give the direction as an angle (in degrees) south of east.

Explanation / Answer

initial speed of ball 1 = v1 = 1.66 m/s (east) initial speed of ball 2 = v2 = 0.81 m/s (west) angle of second ball after collision 2 =90o angle of first ball after collision = 1 mass of first ball = m1 mass of second ball = m2 final speed of first ball = v1' final speed of second ball = v2' = 1.44 m/s According to conservation of momentum we have Along the x-direction      m1v1 - m2v2 = m1v1'cos1 Given that   m1 = m2 ==> v1 - v2 = v1'cos1 = 1.66 - 0.80 =0.85    .....1 Along the y-direction        0 = v1'sin1 -v2'sin90 ==> v1'sin1 = v2' =1.44           ........2 From 1 and 2 we get              v1'sin1 / v1' cos1 = 1.44 / 0.85       ==> tan1 = 1.694       ==> 1 =59.44o ( north of east)                   = 300.5o (south of east) Therefore        v1' = 1.44 / sin1 =1.672 m/s

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