Question

# one billiard ball is shot easy at 2.5 m/s. A second, identical billiard ball is shot...

one billiard ball is shot easy at 2.5 m/s. A second, identical billiard ball is shot west at 1.2m/s. The balls have a glancing collision, not a head-on collision, deflecting the sencond ball by 90 degrees and sending it north at 1.34m/s.
a) what is the speed of the first ball after the collision?
b)what is the direction of the first ball after the collision?

Consider a E-N coordinate system.

u1 = (2.5 m/s) E

u2 = (-1.2 m/s) E

v2 = (1.34 m/s) N

Let v1 be the velocity of second ball after collision and m be the mass of each ball.

Linear momentum of the system is conserved. So,

mu1 + mu2 = mv1 + mv2

=> u1 + u2 = v1 + v2

=> 2.5 E - 1.2 E = v1 + 1.34 N

=> v1 = 1.3 E - 1.34 N

(a) Final speed of first ball, |v1| = [(1.3)2 + (-1.34)2]1/2 = 1.87 m/s

(b) Angle that the final velocity of first ball makes with the horizontal, θ = tan-1(-1.34/1.3) = -45.87o

So, the direction of first ball after collision is 45.87 degrees south of east.

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