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A cue ball initially moving at 3.4 m/s strikes a stationary eight ball of the same size and mass. After the collision, the cue ball’s final speed is 0.94 m/s at an angle of θ with respect to its original line of motion?Find the eight ball’s speed after the col- lision. Assume an elastic collision (ignoring friction and rotational motion).
Answer in units of m/s.

Respuesta :

Answer:

speed of eight ball speed after the collision is 3.27 m/s

Explanation:

given data

initially moving v1i = 3.4 m/s

final speed is v1f = 0.94 m/s

angle = θ w.r.t. original line of motion

solution

we assume elastic collision

so here using conservation of energy

initial kinetic energy = final kinetic energy .............1

before collision kinetic energy = 0.5 × m× (v1i)²

and

after collision kinetic energy =  0.5 × m× (v1f)²  + 0.5 × m× (v2f)²

put in equation 1

0.5 × m× (v1i)² =  0.5 × m× (v1f)²  + 0.5 × m× (v2f)²

(v2f)² = (v1i)² - (v1f)²

(v2f)² = 3.4² - 0.94²

(v2f)² = 10.68

taking the square root both

v2f = 3.27 m/s

speed of eight ball speed after the collision is 3.27 m/s

okpalawalter8

The eight ball speed after the collision is mathematically given as

v_2f = 3.27 m/s

The eight ball speed after the collision

Generally the equation for energy before and after collision is mathematically given as

1/2 × m× (v1i)^2 =  1/2 × m× (v1f)^2  + 1/2 × m× (v2f)^2

Therefore

(v_2f)^2 = 3.4^2 - 0.94^2

v_2f = 3.27 m/s

Therefore,  eight ball speed after the collision is

v_2f = 3.27 m/s

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