Abilliard ball moving at 6.00 m/s strikes a stationary ball of the same mass. after the collision, the first ball moves at 5.12 m/s at an angle of 31.5° with respect to the original line of motion. assuming an elastic collision (and ignoring friction and rotational motion), find the struck ball's velocity after the collision. magnitude m/s direction ° counter-clockwise from the original direction of motion

Magnitude : 3.14 m/s

Direction : 301.31 deg

Explanation:

Assuming that the ball moves in positive x-direction initially.

= mass of each ball

= initial velocity of first ball before collision =

= initial velocity of second ball before collision =

= final velocity of first ball after collision =

= final velocity of second ball after collision = ?

Using conservation of momentum

+ =   +

() + () = () +

= () - ()

=

magnitude of final velocity of second ball is given as

magnitude = sqrt((1.63)² + (- 2.68)²) = 3.14 m/s

Direction is given as

θ = 360 + tan⁻¹(- 2.68/1.63)

θ = 301.31 deg

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s