A.2 Forces and momentum SL Paper 2- IBDP Physics 2025- Exam Style Questions

IBDP Physics SL 2025 – A.2 Forces and momentum SL Paper 2 Exam Style Questions

IBDP Physics 2025 SL Paper 2 – All Chapters

Topic: A.2 Forces and momentum SL Paper 2

Forces, Free-Body Diagrams, Newton’s Laws, Friction, Equilibrium, Momentum, Impulse, Collisions, Circular Motion

Question-A.2 Forces and momentum SL Paper 2

A ball of mass 0.800kg is attached to a string. The distance to the centre of the mass of the ball from the point of support is 95.0cm. The ball is released from rest when the string is horizontal. When the string becomes vertical the ball collides with a block of mass 2.40kg that is at rest on a horizontal surface.

(a) Just before the collision of the ball with the block,
(i) draw a free-body diagram for the ball. [2]

(ii) show that the speed of the ball is about $4.3 \mathrm{~ms}^{-1}$. [1]

(iii) determine the tension in the string. [2]

(b) After the collision, the ball rebounds and the block moves with speed 2.16ms-1.

(i) Show that the collision is elastic. [4]

(ii) Calculate the maximum height risen by the centre of the ball. [2]

(c) The coefficient of dynamic friction between the block and the rough surface is 0.400.
Estimate the distance travelled by the block on the rough surface until it stops. [3]

▶️Answer/Explanation

Ans:

a(i) Tension upwards, weight downwards $\checkmark$ Tension is clearly longer than weight $\checkmark$

( ii )$v=\sqrt{2 \times 9.81 \times 0.95}$ OR $=4.32 \mathrm{kms}^{-1} n$

iii $\begin{aligned} & T-m g=F_{\text {net }} O R \quad T-m g=\frac{m v^2}{r} \checkmark \\ & T \ll=0.800 \times 9.81+\frac{0.800 \times 4.317^2}{0.95} \otimes=23.5 \ll \mathrm{Nw}\end{aligned}$

b i Use of conservation of momentum.
Rebound speed $=2.16 \propto \mathrm{m} \mathrm{s}^{-1}$ w
Calculation of initial KE $=* \frac{1}{2} \times 0.800 \times 4.317^2 n=7.46 * \mathrm{~J}$
Calculation of final KE $=\propto \frac{1}{2} \times 0.800 \times 2.16^2+\frac{1}{2} \times 2.40 \times 2.16^2 n=7.46 \propto \mathrm{J} »$ shence elastic”

b ii ALTERNATIVE 1
Rebound speed is halved so energy less by a factor of $4 \checkmark$ Hence height is $\frac{95}{4}=23.8 \% \mathrm{~cm} \otimes$
ALTERNATIVE 2
Use of conservation of energy / $\frac{1}{2} \times 0.800 \times 2.16^2=0.800 \times 9.8 \times h$ OR

Use of proper kinematics equation (e.g. $0=2.16^2-2 \times 9.8 \times h$ ) $h=23.8 * \mathrm{~cm} \otimes$

c.ALTERNATIVE 1
Frictional force is $f \ll=0.400 \times 2.40 \times 9.81 »=9.42 * N \cdots v$
$9.42 \times d=\frac{1}{2} \times 2.40 \times 2.16^2$ OR $d=\frac{5.5987}{9.42}$
$d=0.594 \ll \mathrm{m} » \checkmark$
ALTERNATIVE 2
$
a=\ll \frac{f}{m}=\mu g=0.4 \times 9.81=» 3.924 \ll \mathrm{m} \mathrm{s}^{-2}{ }_w \checkmark
$
Proper use of kinematics equation(s) to determine
$
d=0.594 \ll \mathrm{m} »
$

Question

Impulse and momentum

The diagram shows an arrangement used to test golf club heads.

The shaft of a club is pivoted and the centre of mass of the club head is raised by a height h before being released. On reaching the vertical position the club head strikes the ball.

a. (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.

[4]

b. 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. [2]

c. 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. [5]

Answer/Explanation

Markscheme

a. (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.

b. 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 );

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.

Examiners report

a. (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.

b. 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.

Question

This question is in two parts. Part 1 is about power and efficiency. Part 2 is about electrical resistance.

Part 1 Power and efficiency

A bus is travelling at a constant speed of 6.2 m s^–1 along a section of road that is inclined at an angle of 6.0° to the horizontal.

a. (i) The bus is represented by the black dot shown below. Draw a labelled sketch to represent the forces acting on the bus.

(ii) State the value of the rate of change of momentum of the bus. [5]

b. The total output power of the engine of the bus is 70kW and the efficiency of the engine is 35%. Calculate the input power to the engine. [2]

c. The mass of the bus is 8.5×10³ kg. Determine the rate of increase of gravitational potential energy of the bus. [3]

d. Using your answer to (c) and the data in (b), estimate the magnitude of the resistive forces acting on the bus. [3]

The engine of the bus suddenly stops working.

(i) Determine the magnitude of the net force opposing the motion of the bus at the instant at which the engine stops.

(ii) Discuss, with reference to the air resistance, the change in the net force as the bus slows down. [4]

Answer/Explanation

Markscheme

a. (i)

identification of normal reaction/N and weight/W;
identification of friction and driving force;
correct directions of all four forces;
correct relative lengths; { (friction ≅ driving force and N ≅ W but N must not be longer than W) (judge by eye)

(ii) zero;

input power =$\frac{{{\rm{output power}}}}{{{\rm{efficiency}}}} = \frac{{70}}{{0.35}}$;
= 200 kW;
Award [2] for a bald correct answer.

height gained in 1s=(6.2 sin 6=) 0.648(m);
rate of change of PE=8.5×10³×9.81×0.648;
=5.4×10⁴W;

d. power used to overcome friction=(7×10⁴−5.4×10⁴=)1.6×10⁴(W); {(allow ECF from (c))

$F = \left( {\frac{p}{v} = } \right)\frac{{1.6 \times {{10}^4}}}{{6.2}}$;
=2.6kN;

(i) component of weight down slope = 8.5×10³×9.81 sin 6;
net force=2.6×10³+8.5×10³×9.81 sin 6
=11kN;
Watch for ECF from (d).

(ii) air resistance decreases as speed drops;
so net force decreases;

Examiners report

a. (i) Diagrams were poorly presented and ill-thought. 4 marks were assigned to this and candidates should have given much more care to it. Marks were given for appropriate descriptions, directions and lengths of the vectors. In particular, candidates should recognize that the term “acceleration” will not do for a driving force, and that “normal” simply implies “at 90°”. The essential point about the upwards force from the surface is that it is a reaction force.

(ii) About half the candidates realized that the momentum change was zero as the velocity was constant.

b. The efficiency calculation was well done by many.

c. This question produced a mixed response varying from excellent fully-explained solutions to incoherent attempts with an incompetent inclusion of components or attempts that focussed on the change in the kinetic energy.

d. Many recognized that the way to estimate the forces was to access the net rate of change of energy and divide this by the speed, but there were two hurdles here: a determination of the correct net power and the correct speed. Very many failed at one or both of these and thus failed to provide a correct answer.