Compare vector and scalar quantities - Leaving Cert Physics - Question 6 - 2014

1. Apparatus and Arrangement: Use weights and pulleys.
2. Procedure and Measurements: Adjust and read the forces carefully using a scale.
3. Observation and Result: Record the resultant force and verify through calculation.

The horizontal force applied by the golfer is calculated as: $F_{horizontal} = 277 ext{ N} \times \cos(24.53^{\circ}) \approx 252 ext{ N}$

The total weight acting downward is 115 N, with a friction force opposing the motion of 252 N. The net force is calculated as: $\text{Net Force} = F_{horizontal} - F_{friction} = 252 ext{ N} - 252 ext{ N} = 0 ext{ N}$

The net force of 0 N indicates that the golfer is moving at a constant speed; there is no acceleration.

According to Newton’s second law:

1. The force is proportional to acceleration: $F \propto (m \cdot u_{y})$
2. Rearranging gives: $F = ma$ where $k = 1$ (by definition of Newton’s law).

Using the equation: <br> $F = ma \implies a = \frac{F}{m}$<br> Convert mass of the ball to kg: $45 ext{ g} = 0.045 ext{ kg}$. Calculate the acceleration as:<br> $F = 5.3 ext{ kN} = 5300 ext{ N}$ <br> Thus, $a = \frac{5300 ext{ N}}{0.045 ext{ kg}} \approx 117777.78 ext{ m/s}^2$. <br> The speed as it leaves the club is: $v = at = 117777.78 ext{ m/s}^2 \times 0.54 ext{ ms} \approx 63.6 ext{ m/s}$

Using the vertical motion equation under gravity: $u_y = 16.6 ext{ m/s}$ The height is given by: $h = \frac{u_y^2}{2g}$ Now substituting: $h = \frac{(16.6 ext{ m/s})^2}{2 \times 9.8 ext{ m/s}^2} \approx 16.38 ext{ m}$