Gnarls B.

# Angle at which the ball moves after the collision

A ball moving at a speed of 6.5 m/s strikes a stationary ball of the same mass. The two balls undergo an elastic collision. After the collision, the first ball moves at an angle of 37º relative to the original line of motion.

Find the angle at which the second ball moves after the collision.

How do you do solve this? Using the sine or cosine law gives me two unknowns...

By:

Gnarls B.

Can you explain why same mass elastic collisions = 90º exit velocity?
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02/25/18

Richard P.

tutor
Let p0 be the initial momentum,  p1 be the final momentum of the first ball and p2 be the final momentum of the second ball..   These are vector quantities.  The conservation of momentum can be written as p0= p1+ p2  .  This is a vector equation.
However, it can be made into a scalar equation by squaring both sides.   The result is  p02  = p12 + p22 + 2 p1 • p2  ,  where p1 • p2  is the scalar dot product of p1 and p2 .

Kinetic energy can be expressed as  p2/2m.   The conservation of kinetic energy can be written as

p02 / 2m = p12 /2m + p22 /2m.

The two conservation equations cannot both hold unless p1 • p2 =0.    This condition means that the vectors  p1 and p2  are perpendicular.  This is the same as saying that the angle between them is 90 degrees.

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02/26/18

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