CAIE A-Level Physics 3 Dynamics
Practise applying Newton’s laws, momentum and force diagrams to mechanics, and analysing friction, drag, gravity and terminal velocity.
- Syllabus
- 2028–2030
- Course
- Physics 9702
- Level
- AS
Practise applying Newton’s laws, momentum and force diagrams to mechanics, and analysing friction, drag, gravity and terminal velocity.
State what is meant by the mass of a body.
mass is the property (of a body/object) resisting changes in motion
or
mass is the quantity of matter (in a body)
B1
Two blocks travel directly towards each other along a horizontal, frictionless surface. The blocks collide, as illustrated in Fig. 3.1.

Fig. 3.1
Block A has mass 3 M and block B has mass M.
Before the collision, block A moves to the right with speed 0.40 ms−1 and block B moves to the left with speed 0.25 ms−1.
After the collision, block A moves to the right with speed 0.20 m s−1 and block B moves to the right with speed v.
Use Newton's third law to explain why, during the collision, the change in momentum of block A is equal and opposite to the change in momentum of block B.
force on A (by B) equal and opposite to force on B (by A) or both A and B exert equal and opposite forces on each other
B1
force is rate of change of momentum and time (of contact) is same
B1
Determine speed v.
p=m v or 3M×0.40 or M×0.25 or 3M×0.2 or M v
C1
(3M×0.40)−(M×0.25)=(3M×0.2)+Mv
C1
v=(3×0.40)−0.25−(3×0.2)=0.35 m s−1
A1
Calculate, for the blocks,
1. the relative speed of approach,
relative speed of approach = ms−1
2. the relative speed of separation.
relative speed of separation = ms−1
1. relative speed of approach =0.40+0.25
=0.65 ms−1
A1
2. relative speed of separation =0.35-0.20
=0.15 ms−1
A1
Use your answers in (b)(iii) to state and explain whether the collision is elastic or inelastic.
(relative) speed of separation not equal to/less than (relative) speed of approach or answers (to (b)(iii) are) not equal and so inelastic collision
B1
A cylinder is suspended from the end of a string. The cylinder is stationary in water with the axis of the cylinder vertical, as shown in Fig. 2.1.

Fig. 2.1 (not to scale)
The cylinder has weight 0.84 N , height h and a circular cross-section of diameter 0.031 m . The density of the water is 1.0×103 kg m−3. The difference between the pressures on the top and bottom faces of the cylinder is 520 Pa .
Calculate the tension T in the string.
T=0.84−0.39=0.45 N
A1
The cylinder in (b) is released from the string at time t=4.0 s. The cylinder falls, from rest, vertically downwards through the water. Assume that the upthrust acting on the cylinder remains constant as it falls.
State the name of the force that acts on the cylinder when it is moving and does not act on the cylinder when it is stationary.
viscous (force)
B1
State and explain the variation, if any, of the acceleration of the cylinder as it falls downwards through the water.
viscous force increases (with speed/time/depth)
B1
(so) acceleration decreases
B1
State the principle of conservation of momentum.
sum/total momentum (of system of bodies) is constant
or
sum/total momentum before = sum/total momentum after
M1
for an isolated system/no (resultant) external force
A1
Ball A moves with speed v along a horizontal frictionless surface towards a stationary ball B, as shown in Fig. 3.1.

Fig. 3.1
Ball A has mass 4.0 kg and ball B has mass 12 kg .
The balls collide and then move apart as shown in Fig. 3.2.
Ball A has velocity 6.0 ms−1 at an angle of θ to the direction of its initial path.
Ball B has velocity 3.5 ms−1 at an angle of 30∘ to the direction of the initial path of ball A.
By considering the components of momentum at right-angles to the direction of the initial path of ball A, calculate θ.
p=m v
C1
(4.0×6.0×sinθ)−(12×3.5×sin30∘)=0
or
(mAvA×sinθ)−(mBvB×sin30∘)=0
M1
θ=61∘
A1
Use your answer in (i) to show that the initial speed v of ball A is 12 ms−1. Explain your working.
shows the horizontal momentum component of ball A or of ball B as ( 4.0×6.0×cosθ ) or ( 12×3.5×cos30∘ )
C1
(4.0×6.0×cos61∘)+(12×3.5×cos30∘)=4.0v so v=12( ms−1)
A1
By calculation of kinetic energies, state and explain whether the collision is elastic or inelastic.
initial EK(=1/2×4.0×122)=290(288)(J)
M1
final EK(=1/2×4.0×6.02+1/2×12×3.52)=150(145.5)(J)
M1
(initial EK> final EK ) so inelastic [both M1 marks required to award this mark]
A1
A child on a sledge slides down a steep hill and then travels in a straight line up an ice-covered slope, as illustrated in Fig. 3.1.

Fig. 3.1 (not to scale)
The sledge passes point A with speed 18 ms−1 at time t=0 and then comes to rest at point B. The child applies a brake to the sledge at point B. The brake does not keep the sledge stationary and it immediately slides back down the slope towards A .
The variation with time t of the velocity v of the sledge from t=0 to t=24 s is shown in Fig. 3.2.

Fig. 3.2
The child and sledge have a total mass of 70 kg . The component of the total weight of the child and sledge that acts down the slope is 80 N .
Determine
the frictional force on the sledge as it moves from B towards A,
F=70×0.50(=35)
C1
frictional force =80-35
=45 N
A1
the angle θ of the slope to the horizontal.
sinθ=80/(70×9.81)
C1
θ=6.7∘
A1