IB Physics HL A 2 Forces and Momentum Questions

Practise IB Physics A.2 by resolving forces in diagrams, applying force models, tracing momentum changes in collisions and explaining the forces needed for circular motion.

Exam points

Apply Newton’s first and second laws to resolved force diagrams for equilibrium or acceleration in linked systems.
Identify Newton’s third-law interaction pairs and distinguish them from balanced forces on one body.
Use Ff ≤ μsFN or Ff = μdFN to calculate friction or the onset of slipping.
Apply Hooke’s law, F = -kx, to calculate spring force from extension and determine its restoring direction.
Combine buoyancy and Stokes drag to calculate a sphere’s terminal speed in a fluid.
Identify fields as the model for action at a distance and determine force interactions for charged particles.
Apply momentum conservation when external resultant force is zero to calculate recoil or post-collision velocity.
Use impulse, J = FΔt = Δp, and force–time data to calculate momentum change or average force.
Compare elastic and inelastic collisions using momentum and kinetic-energy changes.
Use a_c = v²/r and F_c = mv²/r, linking v = ωr or v = 2πr/T in circular motion.

Covered syllabus points

A.2.1—Newton’s three laws of motion A.2.2—Forces as interactions between bodies A.2.3—Free-body diagrams A.2.4—Resultant force from diagrams A.2.5—Contact forces A.2.6—Normal force A.2.7—Frictional force A.2.8—Tension A.2.9—Hooke’s law restoring force A.2.10—Viscous drag A.2.11—Buoyancy A.2.12—Field forces A.2.13—Weight A.2.14—Electric force Fe A.2.15—Magnetic force Fm A.2.16—Linear momentum conservation A.2.17—Impulse A.2.18—Impulse-momentum change A.2.19—Newton’s second law forms A.2.20—Elastic and inelastic collisions A.2.21—Explosions A.2.22—Collision energy A.2.23—Centripetal acceleration A.2.24—Centripetal force A.2.25—Direction change in circular motion A.2.26—Angular and linear speed

[Maximum number: 7]

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 $12^{\circ}$ 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.

(a)

On the diagram, draw and label one other force acting on the skier.

[ 1 ]

arrow vertically downwards labelled weight/ W / mg / gravitational
force $/ \mathrm{F}_{\mathrm{g}} / \mathrm{F}_{\text {gravitational }} /$ force of gravity;
(judge by eye)

Do not allow "gravity".

(b)

Calculate the magnitude of the normal reaction acting on the skier.

[ 2 ]

$\quad(N=) m g \cos \theta /$ correct substitution;
$\left(=73 \times 9.81 \times \cos 12^{\circ}=\right) 700 \mathrm{~N}$ ;

(c)

The total frictional force acting is 65 N . Determine the tension in the rope.

[ 2 ]

tension = frictional force + component of weight parallel to slope /
tension $=65+m g \sin \theta$ ;
$214 / 210 \mathrm{~N}$ ;

(d)

Explain, using Newton's first law of motion, why the resultant force on the skier must be zero.

[ 2 ]

[Maximum number: 15]

A girl on a sledge is moving down a snow slope at a uniform speed.

(a)

Draw the free-body diagram for the sledge at the position shown on the snow slope.

[ 2 ]

arrow vertically downwards labelled weight/gravitational force $F_{\mathrm g}$ ;
arrow perpendicular to the snow slope labelled reaction/normal contact force $F_{\mathrm N}$ ;
friction force acting up the slope

(b)

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.

[ 3 ]

(c)

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 $4.2 \mathrm{~m} \mathrm{~s}^{-1}$ . 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.

[ 2 ]

mention of conservation of momentum

OR

$5.5\times4.2=(55+5.5)v$ $v=0.38\ \mathrm{m\,s^{-1}}$

(d)

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.

[ 3 ]

same change in momentum/impulse;
the time taken to stop would be greater with the snow

$F=\frac{\Delta p}{\Delta t}$ , therefore F is smaller with the snow

OR

force is proportional to rate of change of momentum, therefore F is smaller with the snow

(e)

The sledge, without the girl on it, now travels up a snow slope that makes an angle of $6.5^{\circ}$ to the horizontal. At the start of the slope, the speed of the sledge is $4.2 \mathrm{~m} \mathrm{~s}^{-1}$ . The coefficient of dynamic friction of the sledge on the snow is 0.11 .

[ 3 ]

(i)

Show that the acceleration of the sledge is about $-2 \mathrm{~m} \mathrm{~s}^{-2}$ .

[ 3 ]

friction force down slope $=\mu mg\cos(6.5^\circ)=5.9\ \mathrm N$

component of weight down slope $=mg\sin(6.5^\circ)=6.1\ \mathrm N$

acceleration $=\frac{12}{5.5}=2.2\ \mathrm{m\,s^{-2}}$

(f)

The coefficient of static friction between the sledge and the snow is 0.14 . Outline, with a calculation, the subsequent motion of the sledge.

[ 2 ]

maximum frictional force $=\mu R=7.5\ \mathrm N$ ;
the sledge will not move because the maximum static friction force is greater than the component of weight down the slope

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question (d)

Question 2

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Question (i)

Question (f)