A tennis player serves a ball of mass 0.058 kg. The ball is initially stationary when thrown up and then struck by the racket. Fig. 1.13 shows the force-tim...

Marking Scheme and Worked Solution

(a) Definition of impulse [1 mark]

Impulse is the product of force and the time for which the force acts. Equivalently, impulse equals the change in momentum of the object. [1]

(b) Impulse from the graph [2 marks]

The impulse equals the area under the force-time graph. The graph is approximately triangular. [1]

\( \text{Impulse} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 0.005 \times 800 = 2.0 \text{ N s} \) [1]

(c) Speed of the ball leaving the racket [2 marks]

The ball starts from rest, so impulse = change in momentum = \( mv - 0 \):

\( mv = 2.0 \) [1]

\( v = \frac{2.0}{0.058} = 34.5 \text{ m/s} \) [1]

(d) Time to reach the service box [2 marks]

\( t = \frac{\text{distance}}{\text{speed}} = \frac{18.0}{34.5} \) [1]

\( t = 0.52 \text{ s} \) [1]

(e) Why the actual time is longer [1 mark]

Air resistance acts on the ball during flight, reducing its speed. A lower average speed means the ball takes longer to cover the same distance. [1]

Key Concept: Impulse links force, time, and momentum. On a force-time graph, the area under the curve equals the impulse (and the change in momentum). For a triangular force profile, this is \( \frac{1}{2} \times F_{\max} \times \Delta t \). The impulse-momentum theorem (\( F \Delta t = \Delta p \)) is equivalent to Newton's second law integrated over time. A short contact time with a large force can produce a very large velocity change, as seen in tennis serves, cricket bat strikes, and similar impacts.