IB Physics SL A 2 Forces and Momentum Questions

Practise IB Physics SL A.2 by analysing force diagrams and field interactions, calculating impulse in collisions, and explaining circular-motion forces.

Exam points

Resolve free-body diagrams to identify forces and apply Newton’s laws to equilibrium or acceleration.
Identify action–reaction pairs on different bodies rather than balancing forces on one body.
Apply static and kinetic friction models to find a coefficient or decide whether a body will slip.
Use F = kΔx with a hanging mass and spring extension to determine the spring constant.
Explain terminal motion in a fluid by balancing drag, buoyancy and weight.
Use buoyancy and weight balance to relate a floating object’s submerged fraction to relative density.
Distinguish electric, gravitational and magnetic field effects on charged particles in different frames.
Apply momentum conservation when external resultant force is zero to calculate post-collision speed.
Use impulse, J = Δp, and force–time data to calculate momentum change or mean impact force.
Compare elastic and inelastic collisions using momentum and kinetic-energy conservation or loss.
Use a_c = v²/r and F_c = mv²/r, linking v = ωr to speed, radius and period 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: 9]

A glider is an aircraft with no engine. To be launched, a glider is uniformly accelerated from rest by a cable pulled by a motor that exerts a horizontal force on the glider throughout the launch.

(a)

The glider and pilot have a total mass of 492 kg . During the acceleration the glider is subject to an average resistive force of 160 N . Determine the average tension in the cable as the glider accelerates.

[ 3 ]

$a=\frac{27}{11}=2.45\,\mathrm{m\,s^{-2}}$ $F-160\,\mathrm{N}=492\,\mathrm{kg}\times2.45\,\mathrm{m\,s^{-2}}$ $F=1370\,\mathrm{N}$

This calculation could also be shown in part (a).

Award [0] for a solution that uses $a=9.81\,\mathrm{m\,s^{-2}}$ .

(b)

The cable is wound onto a cylinder of diameter 1.2 m . Calculate the angular velocity of the cylinder at the instant when the glider has a speed of $27 \mathrm{~m} \mathrm{~s}^{-1}$ . Include an appropriate unit for your answer.

[ 2 ]

\omega=\frac{v}{r}=\frac{27\,\mathrm{m\,s^{-1}}}{0.6\,\mathrm{m}}=45\,\mathrm{rad\,s^{-1}}

Do not accept Hz.

Award [1 max] if the unit is missing.

(c)

After takeoff the cable is released and the unpowered glider moves horizontally at constant speed. The wings of the glider provide a lift force. The diagram shows the lift force acting on the glider and the direction of motion of the glider.

Draw the forces acting on the glider to complete the free-body diagram. The dotted lines show the horizontal and vertical directions.

[ 2 ]

(d)

Explain, using appropriate laws of motion, how the forces acting on the glider maintain it in level flight.

[ 2 ]

[Maximum number: 1]

An elevator (lift) and its load accelerate vertically upwards.

Which statement is correct in this situation?

The net force on the load is zero.

The tension in the cable is equal but opposite to the combined weight of the elevator and its load.

The normal reaction force on the load is equal but opposite to the force on the elevator from the load.

The elevator and its load are in translational equilibrium.

[Maximum number: 7]

A cylindrical cork of height H and cross-sectional area A is floating stationary in water. Its depth below the water surface is D.

(a)

Draw and label the forces acting on the cork.

[ 1 ]

Weight and Buoyancy drawn in proper directions (by eye) and correctly identified

Allow any sensible identification of the forces (e.g. mg for $F_{g}$ )

Do not allow "gravity" for weight.

(b)

Show that

\frac{D}{H}=\frac{\rho_{\mathrm{c}}}{\rho_{\mathrm{w}}}

where $\rho_{\mathrm{c}}$ is the density of the cork and $\rho_{\mathrm{w}}$ is the density of water.

[ 2 ]

Weight $=\rho_{c} \cdot A \cdot H \cdot g$

OR

Buoyancy $=\rho_{w} \cdot A \cdot D \cdot g$ $\rho_{c} \cdot A \cdot H \cdot g=\rho_{w} \cdot A \cdot D \cdot g$

<<algebraic manipulation to show the relationship>>

(c)

An icebreaking ship is designed to withstand a collision with an iceberg, a large partially submerged body of ice freely floating in water. The designers model the shape of the iceberg as a cylinder with an approximate cross-sectional area of $4200 \mathrm{~m}^{2}$ and height above sea level of 32 m .

The following data are available:

\begin{aligned} \rho_{\text {ice }} & =920 \mathrm{~kg} \mathrm{~m}^{-3} \\ \rho_{\text {seawater }} & =1030 \mathrm{~kg} \mathrm{~m}^{-3} \end{aligned}

[ 4 ]

(i)

Show that the mass of the iceberg is about $1.2 \times 10^{9} \mathrm{~kg}$ .

The designers assume that the mass of the ship is about $\frac{1}{40}$ the mass of the iceberg and is moving at $12 \mathrm{~m} \mathrm{~s}^{-1}$ when it collides with the iceberg. They stick together after the collision.

[ 2 ]

$\begin{aligned} \text { «(H -32) /H = 920 / } 1030 \text { so } \text { » } \mathrm{H}=300 \text { 《m》 } \text { OR } \mathrm{D}=268 \text { "m" } \mathrm{m}=920 \times 300 \times 4200 \text { OR } 1.16 \times 10^{9} \text { "kg" } \end{aligned}$

Must see either full substitution or answer to 3 or more significant
figures.

(ii)

Calculate the speed of the ship after the collision.

Ice in a still lake will usually form in a single layer on the surface.

[ 2 ]

$\begin{aligned} \frac{1}{40} m 12=\left(\frac{1}{40} m+m\right) v v=0.29\left\langle\mathrm{~m} \mathrm{~s}^{-1}\right\rangle \end{aligned}$

All question bank results loaded

Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question (d)

Question 2

Question 3

Question (a)

Question (b)

Question (c)

Question (i)

Question (ii)