A golf ball strikes a hard, smooth floor at an angle of 22.6 A golf ball strikes

Question

A golf ball strikes a hard, smooth floor at an angle of 22.6

A golf ball strikes a hard, smooth floor at an angle of 22.6 degree and, as the drawing shows, rebounds at the same angle. The mass of the ball is 0.0729 kg. and its speed is 59.4 m/s just before and after striking the floor. What is the magnitude of the impulse applied to the golf ball by the floor? (Hint Note that only the vertical component of the ball's momentum changes during impact with the floor, and ignore the weight of the ball.)

Explanation / Answer

The magnitude of the impulse is simply the change of momentum.

the inital momentum is -m v sin(theta)

the final momentum is m v sin(theta)

The implus would then be = -m v sin(theta)- m v sin(theta)

substitute the known values where m is 0.0729 kg, v is 59.4 m/s, theta is 22.6 degrees

-0.0729 kg (59.4 m/s) sin(22.6 degrees) - 0.0729 kg (59.4 m/s) sin(22.6 degrees)=3.328 N s

Therefore the impulse would be 3.328 N s

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