Question

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?

Answer 1

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) = -45.87^o

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