IB Physics HL A.2 Forces and Momentum Question Bank
Practise IB Physics HL A.2 by solving force, momentum, impulse and circular-motion problems across the shared evidence set.
- Syllabus
- First assessment 2025
- Course
- Physics HL
- Level
- HL
Practise IB Physics HL A.2 by solving force, momentum, impulse and circular-motion problems across the shared evidence set.
This question is about the forces on a skier.
A skier is pulled up a hill by a rope at a steady velocity. The hill makes an angle of 12∘ with the horizontal. The mass of the skier and skis is 73 kg . The diagram below shows three of the forces acting on the skier.

On the diagram, draw and label one other force acting on the skier.
arrow vertically downwards labelled weight/ W / mg / gravitational
force /Fg/Fgravitational / force of gravity; (judge by eye)
Do not allow "gravity".
Calculate the magnitude of the normal reaction acting on the skier.
(N=)mgcosθ/ correct substitution;
(=73×9.81×cos12∘=)700 N;
The total frictional force acting is 65 N . Determine the tension in the rope.
tension = frictional force + component of weight parallel to slope /
tension =65+mgsinθ;
214/210 N;
Explain, using Newton's first law of motion, why the resultant force on the skier must be zero.
(Newton's first law states that a body remains at rest or moves with) constant velocity/steady speed/uniform motion unless external/net/resultant/unbalanced force acts on it;
clear link that in this case there is constant/steady velocity so no resultant force;
A girl on a sledge is moving down a snow slope at a uniform speed.

Draw the free-body diagram for the sledge at the position shown on the snow slope.
arrow vertically downwards labelled weight «of sledge and/or girl»/W/mg/gravitational force/ Fg/Fgravitational AND arrow perpendicular to the snow slope labelled reaction force/R/normal contact force/N/ FN friction force/F/f acting up slope «perpendicular to reaction force»
Do not allow G/g/"gravity".
Do not award MP1 if a "driving force" is included. Allow components of weight if correctly labelled. Ignore point of application or shape of object.
Ignore "air resistance". Ignore any reference to "push of feet on sledge".
Do not award MP2 for forces on sledge on horizontal ground
The arrows should contact the object
After leaving the snow slope, the girl on the sledge moves over a horizontal region of snow. Explain, with reference to the physical origin of the forces, why the vertical forces on the girl must be in equilibrium as she moves over the horizontal region.
gravitational force/weight from the Earth «downwards» reaction force from the sledge/snow/ground «upwards» no vertical acceleration/remains in contact with the ground/does not move vertically as there is no resultant vertical force
Marking guidance:
Allow naming of forces as in (a)
Allow vertical forces are balanced/equal in magnitude/cancel out
When the sledge is moving on the horizontal region of the snow, the girl jumps off the sledge. The girl has no horizontal velocity after the jump. The velocity of the sledge immediately after the girl jumps off is 4.2 m s−1. The mass of the girl is 55 kg and the mass of the sledge is 5.5 kg . Calculate the speed of the sledge immediately before the girl jumps from it.
mention of conservation of momentum
OR
5.5×4.2=(55+5.5)≪/><
0.38 « ms−1 »
Marking guidance:
Allow p=p′ or other algebraically equivalent statement
Award [0] for answers based on energy
The girl chooses to jump so that she lands on loosely-packed snow rather than frozen ice. Outline why she chooses to land on the snow.
same change in momentum/impulse the time taken «to stop» would be greater «with the snow»
F=ΔtΔp therefore F is smaller «with the snow»
OR
force is proportional to rate of change of momentum therefore F is smaller «with the snow»
Marking guidance:
Allow reverse argument for ice
The sledge, without the girl on it, now travels up a snow slope that makes an angle of 6.5∘ to the horizontal. At the start of the slope, the speed of the sledge is 4.2 m s−1. The coefficient of dynamic friction of the sledge on the snow is 0.11 .
Show that the acceleration of the sledge is about −2 m s−2.
«friction force down slope» =μmgcos(6.5)= «5.9N» «component of weight down slope» =mgsin(6.5) « =6.1 N »
«so a=mF » acceleration =5.512=2.2 « ms−2 »
Marking guidance:
Ignore negative signs Allow use of g=10 ms−2
The coefficient of static friction between the sledge and the snow is 0.14 . Outline, with a calculation, the subsequent motion of the sledge.
calculates a maximum value for the frictional force = « μR= » 7.5 « N » sledge will not move as the maximum static friction force is greater than the component of weight down the slope
Marking guidance:
Allow correct conclusion from incorrect MP1 Allow 7.5 > 6.1 so will not move