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In an oblique collision, the velocity of the object is not along the line of impact. This requires resolving velocities into components parallel and perpendicular to the surface or line of centres.
When a particle hits a smooth wall obliquely:
A ball hits a smooth wall at 30 degrees to the wall with speed 12 m s^(-1). The coefficient of restitution is 0.8. Find the speed and direction after the bounce.
Solution:
Perpendicular component (towards wall): 12 sin 30 = 6 m s^(-1)
Parallel component (along wall): 12 cos 30 = 6 sqrt(3) m s^(-1)
After bounce:
Speed after = sqrt(4.8^2 + (6 sqrt(3))^2) = sqrt(23.04 + 108) = sqrt(131.04) approximately 11.4 m s^(-1)
Angle to wall: tan(beta) = 4.8 / (6 sqrt(3)) = 4.8/10.39 approximately 0.462
beta = arctan(0.462) approximately 24.8 degrees
If the ball hits the wall at angle alpha to the wall surface and rebounds at angle beta:
tan beta = e tan alpha
This follows because:
So tan(beta) / tan(alpha) = (e * perpendicular) / perpendicular * (parallel / parallel) = e.
When e = 1: beta = alpha (angle of incidence = angle of reflection, like light).
A ball strikes a wall at 45 degrees with e = 0.5. Find the angle of rebound.
tan beta = 0.5 * tan 45 = 0.5 * 1 = 0.5
beta = arctan(0.5) approximately 26.6 degrees to the wall.
When two smooth spheres collide obliquely:
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