Abilliard ball moving at 6.00 m/s strikes a stationary ball of the same mass. after the collision, the first ball moves at 5.39 m/s at an angle of 26.0° 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.

The final velocity of the second billiard ball is 11.1 m/s.

The given parameters;

Apply the principle of conservation of linear momentum for elastic collision.

m₁u₁ + m₂u₂ = m₁v₁  + m₂v₂

Apply one dimensional velocity in x-direction;

Apply one dimensional velocity in y-direction

The resultant of the second ball final velocity is calculated as follows;

Thus, the final velocity of the second billiard ball is 11.1 m/s.

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Explanation:

In the initial direction of motion we can use momentum conservation

so here we have

now by momentum conservation along perpendicular direction we can use as

So the magnitude of the final speed of the stuck ball is given as

direction of motion given as

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