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A ball moving with speed v strikes a smooth wall at an angle of 60 degrees to the normal. The component of velocity along the wall is preserved during the collision. If the normal component of velocity after impact is v', and the coefficient of restitution is e = 1/2, find the speed of the ball after collision.
- v * sqrt(3) / 2
- v * sqrt(7) / 4
- v * sqrt(5) / 4
- v / 2
Correct answer: v * sqrt(7) / 4
Solution
The ball strikes the wall at 60 degrees to the normal. The angle to the wall surface is 30 degrees, so the angle to the normal is 60 degrees. Tangential speed (along wall) = v*cos(60 deg) = v/2 (preserved). Normal speed before impact = v*sin(60 deg) = v*sqrt(3)/2. With e = 1/2, normal speed after = e * v*sqrt(3)/2 = v*sqrt(3)/4. Final speed = sqrt((v/2)² + (v*sqrt(3)/4)²) = sqrt(v²/4 + 3v²/16) = sqrt(4v²/16 + 3v²/16) = sqrt(7v²/16) = v*sqrt(7)/4.
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