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.

u_{1} = (2.5 m/s) E

u_{2} = (-1.2 m/s) E

v_{2} = (1.34 m/s) N

Let v_{1} be the velocity of second ball after collision
and m be the mass of each ball.

Linear momentum of the system is conserved. So,

mu_{1} + mu_{2} =
mv_{1} + mv_{2}

=> u_{1} +
u_{2} = v_{1} + v_{2}

=> 2.5 E - 1.2 E =
v_{1} + 1.34 N

=> v_{1} = 1.3 E -
1.34 N

(a) Final speed of first ball,
|v_{1}| = [(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.**

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