Question

Billiard ball collision in 2-D. Billiard ball A moving with speed v_A= 6.9-m/s in the +x direction strikes an equal-mass ball B initially at rest.

The two balls are observed to move off at some unknown angle to the x-axis, ball A above the x-axis and ball B below.

If v'A = 2.86-m/s at what angle did ball A go above the x-axis?

Answer 1

let v = 6.9 m/s
v1 = 2.86 m/s
let v2 is the velocity of ball B after the collision.

let m is the mass of each ball.

Apply conservation of energy

(1/2)*m*v^2 = (1/2)*m*v1^2 + (1/2)*m*v2^2

v^2 = v1^2 + v2^2

v2 = sqrt(v^2 - v1^2)

= sqrt(6.9^2 - 2.86^2)

= 6.28 m/s

let A is the angle made by ball A above +x axis

let B is the angle made by ball B below +x axis

we know, A + B = 90 degrees

Apply conservation of momentum in y-direction

initial momentum in y-direction = final momentum in y-direction

0 = m*v1*sin(A) - m*v2*sin(B)

v1*sin(A) = v2*sin(B)

v1*sin(A) = v2*sin(90- A)

v1*sin(A) = v2*cos(A)

sin(A)/cos(A) = v2/v1

tan(A) = 6.28/2.86

A = tan^-1(6.28/2.86)

= 65.5 degrees <<<<<<<<<---------------Answer