(i) Describe the energy changes that take place in the club head from the instant the club is released until the club head and the ball separate.

(ii) Calculate the maximum speed of the club head achievable when h = 0.85 m.

The diagram shows the deformation of a golf ball and club head as they collide during a test.

Explain how increasing the deformation of the club head may be expected to increase the speed at which the ball leaves the club.

In a different experimental arrangement, the club head is in contact with the ball for a time of 220 μs. The club head has mass 0.17 kg and the ball has mass 0.045 kg. At the moment of contact the ball is at rest and the club head is moving with a speed of 38 ms^–1. The ball moves off with an initial speed of 63 ms^–1.

(i) Calculate the average force acting on the ball while the club head is in contact with the ball.

(ii) State the average force acting on the club head while it is in contact with the ball.

(iii) Calculate the speed of the club head at the instant that it loses contact with the ball.

(i) (gravitational) potential energy (of club head) goes to kinetic energy (of club head);

some kinetic energy of club head goes to internal energy of club head/kinetic energy of ball; 

(ii) equating mgh to \(\frac{1}{2}\)mv² ;  

v=4.1(ms^-1);

Award [0] for answers using equation of motion – not uniform acceleration. 

deformation prolongs the contact time;

increased impulse => bigger change of momentum/velocity;
or
(club head) stores (elastic) potential energy on compression;

this energy is passed to the ball;

(i) any value of \(\frac{{{\rm{mass}} \times {\rm{velocity}}}}{{{\rm{time}}}}\);

1.3 x 10⁴ (N);

(ii)-1.3 x 10^-4 (N);

Accept statement that force is in the opposite direction to (c)(i).

Allow the negative of any value given in (c)(i).

(iii) clear use of conservation of momentum / impulse = change of momentum;

21(ms^-1 );

or

a=\(\left( {\frac{{\rm{F}}}{{\rm{m}}} = \frac{{ - 13000}}{{0.17}} = } \right)\left(  -  \right)76500\left( {{\rm{m}}{{\rm{s}}^{ - 1}}} \right)\);
v=(u+at=38-76500 x 0.00022=) 21(ms^-1);

Award [2] for a bald correct answer.

(i) Nearly all candidates gained a mark for recognising the change from kinetic to potential energy in this part.  Fewer recognised that the club head would not transfer all of its energy to the ball and therefore retained a significant amount of energy. 

(ii) This part was well done by many.

 A minority of candidates became bogged down by the deformation of the ball and club head idea and ventured into elastic potential energy ideas. This had a successful outcome in many cases when there was discussion of the compression providing further kinetic energy to the ball on recovering its shape. The most straightforward solution was to use to principle of impulse being equal to the change in momentum (as shown in the question heading) and simply to recognise that an increased contact time would be expected to give a greater change of momentum for a constant force.

(i) This was well done with the only real problem being deciding which was the speed change of the ball. 

(ii) Less candidates than anticipated recognised that the force on the club head was equal and opposite to that acting on the ball (applying Newton’s third law of motion).

(iii) Most made a good attempt at calculating the speed of the club head.