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A billiard ball moving at 2.5 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 2.17 m/s, at an angle of 30.0° 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.

Sagot :

First, let's draw a diagram

In order to solve this problem, we will need to consider the conservation of momentum

Initially we have a ball with mass m moving at 2.5 m/s with 0 degrees from the axis. After the collision, we have 2 balls, each with the same amount of energy from before the collision since the ball had an elastic collision.

Currently, we know that initally there was no vertical component of velocity, which means that the struck ball has an equal and opposite vertical component as the ball that was hit.

2.17sin30 = 1.085 m/s in the vertical direction

Now we need to find the horizontal component.

Since the struck ball has a velocity of 2.5 m/s, we know in the end, both ball's combined horizontal velocities should add up to 2.5 m/s

2.17 cos 30 = 1.879 m/s

2.5 - 1.879 = .621 m/s in the horizontal direction.

now we can combine both of these directions to get the final velocity

vf = sqrt(.621^2+1.085^2) = 1.25 m/s

View image HennleyA380710
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