Question

A golf ball strikes a hard, smooth floor at an angle of 26.3 ° and, as the drawing shows, rebounds at the same angle. The mass of the ball is 0.0457 kg, and its speed is 58.7 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.)

Answer 1

A golf ball strikes a hard, smooth floor at an angle of $26.3{}^{\circ}$ and, as the drawing shows, rebounds at the same angle. The mass of the ball is $0.0457\mathrm{kg}$, and its speed is $58.7\mathrm{m/s}$ 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.)

Solution

Impulse = Change in momentum

Resolve the velocity vector into its rectangular components. Now, since speed before and after striking the floor remains the same, hence the horizontal component ($v \cos \theta$) doesn't change. Hence, only the vertical components come into play.

Therefore, Initial momentum = $m v \sin \theta$.

And, final momentum = $-m v \sin \theta$.

Thus, Impulse = $-m v \sin \theta - m v \sin \theta = -2 m v \sin \theta$.

The magnitude of the impulse applied to the golf ball by the floor

$$\Delta P = 2 m v \sin \theta = 2 * 0.0457 * 58.7 * \sin 26.3{}^{\circ} = 2.38 \frac{\mathrm{kg} * \mathrm{m}}{\mathrm{s}}.$$

Answer: $2.38 \frac{\mathrm{kg} * \mathrm{m}}{\mathrm{s}}$.