Question

One billiard ball is shot east at 2.5 m/s . A second, identical billiard ball is shot west at 0.80 m/s . The balls have a glancing collision, not a head-on collision, deflecting the second ball by 90∘ and sending it north at 1.34 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 south of east.

Answer 1

he mass of the identical balls m1 = m2 = m

the velocity of the first ball v1 = 2.5i m/s

the velocity of the second ball v2 = -0.80 i m/s

the final velocity of the second ball v2f = 1.34j m/s

Here i ,j are the units vectors along east and north

Now from law of conservation of momentum

m1v1i + m2v2i = m1v1f + m2v2f

along X- axis

m1vx1f = m1v1i + m2v2i

vx1f = v1i + v2i

= 2.5 - 0.80

= 1.7i m/s

along Y-axis

0 = m1vy1f + m2v2f

m1vy1f = -m2v2

then vyf1 = - v2f = -1.34j

then the final speed of the first ball

v1f = sqrt(1.7)^2 + (1.34)^2

v1f = 2.165 m/s

and direction

θ = tan^-1(-1.34/1.7) = 38.24 deg south of east