A billiard ball strikes identical ball and is deflected 45 degrees from original direction. Original ball moves 10m/s, if collision elastic, what's the velocity of ball after collision?
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Hello
If the first ball is deflected by 45°, then is the other ball also deflected by 45°, because there is a 90° angle between the balls after elastic collision, and both balls have identical velocities.
the kinetic energy is conserved so
1/2 mv1^2 = 1/2*mv1'^2 + 1/2*mv2'^2
v1^2 = v1'^2 + v1'^2 = 2v1'^2
50 = v1'2
v1' = √50 m/s <--- velocity of the balls after collision
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also the momentum is conserved:
net momentum in y direction = 0
net momentum in x direction = mv1' *cos45 + mv1' *cos(-45) = mv1
2 v1' *cos 45 = v1
2*√50 * cos 45 = 10 = v1
Regards
If the first ball is deflected by 45°, then is the other ball also deflected by 45°, because there is a 90° angle between the balls after elastic collision, and both balls have identical velocities.
the kinetic energy is conserved so
1/2 mv1^2 = 1/2*mv1'^2 + 1/2*mv2'^2
v1^2 = v1'^2 + v1'^2 = 2v1'^2
50 = v1'2
v1' = √50 m/s <--- velocity of the balls after collision
--------
also the momentum is conserved:
net momentum in y direction = 0
net momentum in x direction = mv1' *cos45 + mv1' *cos(-45) = mv1
2 v1' *cos 45 = v1
2*√50 * cos 45 = 10 = v1
Regards