Velocity problem help?

Question

A billiard ball moving at 5.10 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 4.28 m/s, at an angle of 33.0° with respect to the original line of motion.

(a) Find the velocity (magnitude and direction) of the second ball after collision.

______m/s

______° (with respect to the original line of motion, include the sign of your answer; consider the sign of the first ball's angle)

*please explain

Answer 1

Conservation of Momentum.  The momentum of any two objects before a collision will equal the amount of momentum after the collision.  Because the objects go off in different directions we have to solve a triangle.  Mass isn't an issue so we'll deal solely with the velocities.

Before the collision the first ball had only horizontal momentum, and the second ball had none, therefore the total X momentum (horizontal) is 5.1 m/s.

After the collision the first ball goes 4.28 m/s at an angle of 33°.  Draw a triangle showing hypotenuse of 4.28.  Use sin33*4.28 to find the opposite side representing the vertical velocity of the ball.  This = 2.33 m/s.  Use cos33 * 4.28 to find the adjacent side, or horizontal velocity.  This = 3.59 m/s.

Draw a new triangle with the sum of the horizontal and vertical velocities.  Horizontal = 8.69 m/s and vertical = 2.33 m.  Use tan (opposite/adjacent) to find the angle of deflection.  Tan 2.33/8.69 = 15° with respect to its original horizontal motion.

Do Pythagorean theorem to find the hypotenuse.  2.33 squared + 8.69 squared = h squared.  Square root of h is 8.99 m/s.

The velocity of the second ball after the collision is 8.99 m/s  at 15°.