Worked Solutions
Module 2: Dynamics — Worked Solutions (Preliminary Physics)
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Worked examples for Preliminary Physics Module 2: Dynamics. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
For force problems, draw a free-body diagram and resolve forces along sensible axes. For collisions, check whether momentum (always conserved here) or energy is the right tool.
Example 1 — Newton's second law on a frictionless incline
Question
A $2.5\ \text{kg}$ block is released from rest on a smooth (frictionless) ramp inclined at $30^\circ$ to the horizontal. Taking $g = 9.8\ \text{m s}^{-2}$, find the block's acceleration down the ramp and its speed after sliding $4.0\ \text{m}$ along the surface.
Solution
Only the component of gravity along the ramp drives the motion; the normal force is perpendicular and the ramp is smooth.
Along the ramp: $F = mg\sin\theta = 2.5 \times 9.8 \times \sin 30^\circ = 2.5 \times 9.8 \times 0.5 = 12.25\ \text{N}$.
Newton's second law: $a = \dfrac{F}{m} = \dfrac{12.25}{2.5} = 4.9\ \text{m s}^{-2}$. (Equivalently $a = g\sin\theta$.)
Speed after $4.0\ \text{m}$ from rest: $v^2 = u^2 + 2as = 0 + 2(4.9)(4.0) = 39.2$, so $v = 6.3\ \text{m s}^{-1}$.
Note the mass cancels for the acceleration — don't be surprised it doesn't appear in $a = g\sin\theta$.
The first move is to resolve gravity into components: one along the ramp (which makes the block slide) and one pressing into the ramp (balanced by the normal force). Because the ramp is frictionless, only the along-ramp component matters.
That component is $mg\sin\theta = 2.5 \times 9.8 \times \sin 30^\circ = 12.25\ \text{N}$.
Newton's second law, $F = ma$, then gives $a = \dfrac{12.25}{2.5} = 4.9\ \text{m s}^{-2}$. Notice this equals $g\sin\theta$ — the mass cancels, which is why all objects slide down a smooth ramp at the same rate.
Now use $v^2 = u^2 + 2as$ with $u = 0$: $v^2 = 2 \times 4.9 \times 4.0 = 39.2$, so $v = \sqrt{39.2} = 6.3\ \text{m s}^{-1}$.
We use the equation without time because we're given distance, not duration.
Smooth incline: driving force $= mg\sin\theta$.
- $F = 2.5 \times 9.8 \times 0.5 = 12.25\ \text{N}$
- $a = F/m = 12.25/2.5 = 4.9\ \text{m s}^{-2}$ ($= g\sin\theta$)
- $v^2 = 0 + 2(4.9)(4.0) = 39.2$
- $v = 6.3\ \text{m s}^{-1}$
$a = 4.9\ \text{m s}^{-2}$; $v = 6.3\ \text{m s}^{-1}$.
Where the marks go
- 1 mark: Resolves gravity along the ramp: $F = mg\sin\theta$
- 1 mark: Correct acceleration $a = 4.9\ \text{m s}^{-2}$
- 1 mark: Uses $v^2 = u^2 + 2as$ to get $v = 6.3\ \text{m s}^{-1}$
Key idea
On a smooth incline the net force is $mg\sin\theta$, so $a = g\sin\theta$ — independent of mass.
Example 2 — Conservation of momentum in a collision
Question
A $1200\ \text{kg}$ car travelling east at $20\ \text{m s}^{-1}$ collides with a stationary $800\ \text{kg}$ car. The two cars lock together and move off as one. Find their common velocity immediately after the collision, and determine whether the collision is elastic by comparing the kinetic energy before and after.
Solution
Momentum is conserved in any collision. Take east as positive.
Before: $p = 1200 \times 20 + 800 \times 0 = 24\,000\ \text{kg m s}^{-1}$.
After (combined mass $2000\ \text{kg}$): $24\,000 = 2000\,v \Rightarrow v = 12\ \text{m s}^{-1}$ east.
Kinetic energy before: $\tfrac{1}{2}(1200)(20)^2 = 240\,000\ \text{J}$.
Kinetic energy after: $\tfrac{1}{2}(2000)(12)^2 = 144\,000\ \text{J}$.
Energy dropped from $240\ \text{kJ}$ to $144\ \text{kJ}$, so the collision is inelastic. Cars locking together is the classic sign — never assume KE is conserved.
Whenever two objects collide with no external horizontal force, the total momentum stays the same — that's our anchor. Taking east as positive, the momentum before is all in the moving car:
$p_{\text{before}} = 1200 \times 20 + 800 \times 0 = 24\,000\ \text{kg m s}^{-1}$.
After the crash they're stuck together, so they share a mass of $2000\ \text{kg}$ and one velocity $v$:
$2000\,v = 24\,000 \Rightarrow v = 12\ \text{m s}^{-1}$ east.
Now check the energy. Before: $\tfrac{1}{2}(1200)(20)^2 = 240\,000\ \text{J}$. After: $\tfrac{1}{2}(2000)(12)^2 = 144\,000\ \text{J}$. Since $144\,000\ \text{J} < 240\,000\ \text{J}$, kinetic energy was lost, so the collision is inelastic. That missing $96\,000\ \text{J}$ went into deforming the cars, heat and sound — which is exactly what we'd expect when objects crumple and stick.
East positive. Momentum conserved.
- Before: $p = 1200(20) = 24\,000\ \text{kg m s}^{-1}$
- After: $v = 24\,000 / 2000 = 12\ \text{m s}^{-1}$ east
- $KE_i = \tfrac{1}{2}(1200)(20)^2 = 240\,000\ \text{J}$
- $KE_f = \tfrac{1}{2}(2000)(12)^2 = 144\,000\ \text{J}$
$KE_f < KE_i$ → inelastic; $v = 12\ \text{m s}^{-1}$ east.
Where the marks go
- 1 mark: Applies conservation of momentum with correct before-state
- 1 mark: Correct common velocity $12\ \text{m s}^{-1}$ east
- 1 mark: Correct kinetic energies before ($240\ \text{kJ}$) and after ($144\ \text{kJ}$)
- 1 mark: Concludes inelastic because kinetic energy is not conserved
Key idea
Momentum is conserved in all collisions; kinetic energy is conserved only in elastic ones, so compare KE before and after to classify.