Question

Two billiard balls with the same mass m move in the direction of one another. Ball one travels in the positive x-direction with a speed of V1i, and ball two travels in the negative x-direction with a speed of V2i. The two balls collide elastically, and both balls change direction after collision. If the initial speeds of the balls were v1i=2.0m/s and v2i =1.0m/s, what would be the final speed and direction of ball two, v2f, in m/s?    Thank you so much.

Answer 1

Given,

mass of both the ball are i.e m

Initial speed of ball one, v_1i = 2 m/s    in positive x-direction

Initial speed of ball two, v_2i = 1 m/s    in negative x-direction

After collision both ball change the directions

Let final speed of ball one after collision be v_1f   which is in negative x-direction

Let final speed of ball two after collision be v_2f   which is in positive x-direction

Now,

since collision is elastic, applying conservation of momentum

=> initial momentum = final momentum

=> mv_1i - mv_2i = -mv_1f + mv_2f

=> v_1i - v_2i = -v_1f + v_2f

=> 2 - 1 = -v_1f + v_2f

=> -v_1f + v_2f = 1

=> v_1f = v_2f - 1

Now,

applying conservation of energy

=> initial kinetic energy = final kinetic energy

=> 1/2*m*(v_1i)² + 1/2*m*(v_2i)² = 1/2*m*(v_1f)² + 1/2*m*(v_2f)²

=> (v_1i)² + (v_2i)² = (v_1f)² + (v_2f)²

=> 2² + 1² = (v_1f)² + (v_2f)²

=> (v_1f)² + (v_2f)² = 5

Now put v_1f = v_2f - 1

=> (v_2f - 1)² + (v_2f)² = 5

=> (v_2f)² - 2*(v_2f) + 1 + (v_2f)² = 5

=> 2*(v_2f)² - 2*(v_2f) - 4 = 0

=> (v_2f)² - (v_2f) - 2 = 0

or   v_2f = =

              = ( 1 3) / 2

or v_2f = 2 m/s or v_2f = -1 m/s

Now,

Since final velocity of ball two is in +x-direction,

hence,   v_2f = 2 m/s , i.e two masses interchange their velocities.

final speed of ball two is 2 m/s in +x-direction.