Physics 9702/22 — May/June 2016

Background Concept

For motion with constant acceleration, acceleration is defined by

$$a = \frac{\Delta v}{\Delta t} = \frac{v-u}{t}$$

where $u$ is initial velocity, $v$ is final velocity, and $t$ is the time taken.

Understanding the Question

Between A and B, the toboggan starts from rest and reaches a speed of $36\ \text{m s}^{-1}$ in $19\ \text{s}$, with constant acceleration. The question asks for that acceleration.

Approach

Use the definition of acceleration for constant acceleration:

1. Identify $u$, $v$, and $t$.
2. Substitute into $a = (v-u)/t$.

Step-by-Step Reasoning

• Starts from rest, so $u=0$.
• Final speed at B is $v=36\ \text{m s}^{-1}$.
• Time to reach B is $t=19\ \text{s}$.

$$a = \frac{36-0}{19} = 1.8947\ldots\ \text{m s}^{-2}$$

To 2 significant figures (matching the data),

$$a \approx 1.9\ \text{m s}^{-2}$$

Key Takeaways

• Constant acceleration lets you use simple kinematics relations.
• Always check you have consistent units (here, SI already).

Common Mistakes

• Using $a = v/t$ without stating $u=0$ (works here but can be unsafe in general).
• Rounding too early (keep extra digits until the end).

Things to Be Careful About

• “Speed” is the magnitude of velocity; since motion is along the slope, treating it as 1D is fine.
• Give the unit $\text{m s}^{-2}$ for acceleration.

Background Concept

For constant acceleration, the displacement can be found from the mean velocity:

$$\text{mean velocity} = \frac{u+v}{2}$$

and

$$s = (\text{mean velocity})\, t$$

This works because with uniform acceleration, velocity increases linearly with time.

Understanding the Question

You are asked to show that the distance from A to B is $340\ \text{m}$, using the information already given for motion from rest to $36\ \text{m s}^{-1}$ in $19\ \text{s}$.

Approach

Use

$$s = \frac{u+v}{2}t$$

with $u=0$, $v=36\ \text{m s}^{-1}$, $t=19\ \text{s}$, then round to the stated value $340\ \text{m}$.

Step-by-Step Reasoning

• Since the acceleration is constant, the average speed over the interval is the average of initial and final speeds.

$$\bar{v} = \frac{0 + 36}{2} = 18\ \text{m s}^{-1}$$

• Distance is average speed times time:

$$s = 18 \times 19 = 342\ \text{m}$$

• The question says “show that” it is $340\ \text{m}$, so we round appropriately:

$$342\ \text{m} \approx 340\ \text{m}$$

Key Takeaways

• For uniform acceleration, $s = \frac{u+v}{2}t$ is often the quickest route.
• “Show that” questions usually expect an approximate match after rounding.

Common Mistakes

• Using $s = vt$ with $v=36\ \text{m s}^{-1}$ (that would assume constant velocity, not acceleration).
• Rounding $a$ too much in part (i) and then using $s = \frac{1}{2}at^2$ to get a poor value.

Things to Be Careful About

• Keep at least 3 significant figures in intermediate steps if you use $s = \frac{1}{2}at^2$.
• Ensure you are calculating distance along the slope, not vertical height (height is needed later).

Background Concept

Two key energy expressions are used here:

• Kinetic energy:

$$E_k = \frac{1}{2}mv^2$$

• Gravitational potential energy (near Earth’s surface):

$$E_p = mgh$$

A change in gravitational potential energy depends on the vertical height change $h$, not the distance travelled along the slope.

Understanding the Question

From A to B:

• Mass of man + toboggan: $m = 95\ \text{kg}$.
• Starts from rest: $u=0$.
• Speed at B: $v = 36\ \text{m s}^{-1}$.
• Slope angle: $40^{\circ}$ to the horizontal.
• Distance along the slope A to B (from part (ii)): $s = 340\ \text{m}$.

You must calculate:

1. the increase in kinetic energy from A to B,
2. the change in gravitational potential energy from A to B.

Approach

1. Use $\Delta E_k = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$.
2. Find the vertical drop $h$ using the slope geometry: $h = s\sin\theta$.
3. Then use $\Delta E_p = -mgh$ (negative because the toboggan loses height).

Step-by-Step Reasoning

1) Change in kinetic energy

At A, $u=0$ so initial kinetic energy is zero.

$$\Delta E_k = \frac{1}{2}mv^2 = \frac{1}{2}(95)(36^2)$$

Compute $36^2 = 1296$:

$$\Delta E_k = 0.5 \times 95 \times 1296 = 61560\ \text{J} \approx 6.2 \times 10^{4}\ \text{J}$$

2) Change in potential energy

The distance $340\ \text{m}$ is along the slope; convert to vertical drop using trigonometry:

$$h = s\sin 40^{\circ}$$ $$h = 340\sin 40^{\circ} \approx 340 \times 0.643 \approx 2.19 \times 10^{2}\ \text{m}$$

Now compute the loss of gravitational potential energy:

$$\Delta E_p = -mgh = -(95)(9.81)(2.19 \times 10^{2}) \approx -2.0 \times 10^{5}\ \text{J}$$

Negative sign indicates a decrease in potential energy.

Key Takeaways

• $\Delta E_k$ depends on speed change.
• $\Delta E_p$ depends only on vertical height change, not slope length.
• A downhill motion gives a negative $\Delta E_p$.

Common Mistakes

• Using $h = s\cos 40^{\circ}$ instead of $s\sin 40^{\circ}$.
• Giving $\Delta E_p$ as positive when the object moves downwards.
• Using $s=342\ \text{m}$ in one step and $s=340\ \text{m}$ in another without consistency (either is fine if rounded consistently).

Things to Be Careful About

• Keep enough significant figures during intermediate calculations to avoid rounding errors.
• State energies in joules and use $g = 9.81\ \text{m s}^{-2}$ (or consistent value).

Background Concept

When friction is present, mechanical energy is not conserved, but total energy is. The loss of gravitational potential energy becomes:

• an increase in kinetic energy, and
• energy dissipated as thermal energy / sound due to friction.

Work done against friction is

$$W_{\text{fr}} = F_{\text{fr}}\, s$$

where $s$ is the distance moved along the surface.

Understanding the Question

From A to B, you have already found:

• increase in kinetic energy $\Delta E_k$,
• decrease in potential energy $\Delta E_p$ (negative),
• distance along slope $s = 340\ \text{m}$.

You are asked for the average frictional force between A and B.

Approach

1. Find the energy dissipated by friction:

$$W_{\text{fr}} = \left|\Delta E_p\right| - \Delta E_k$$

1. Convert this work to an average friction force using $F = W/s$.

Step-by-Step Reasoning

Because the toboggan goes down the slope, it loses gravitational potential energy. If there were no friction, all of that loss would become kinetic energy.

But friction is present, so the kinetic energy gained is smaller than the potential energy lost.

• Magnitude of potential energy loss: $\left|\Delta E_p\right|$.
• Kinetic energy gain: $\Delta E_k$.

So the “missing” energy is what friction has removed:

$$W_{\text{fr}} = \left|\Delta E_p\right| - \Delta E_k$$

Substitute values from (iii):

$$W_{\text{fr}} = 2.04 \times 10^{5} - 6.16 \times 10^{4} = 1.42 \times 10^{5}\ \text{J}$$

Then use $W = Fs$:

$$F_{\text{fr}} = \frac{W_{\text{fr}}}{s} = \frac{1.42 \times 10^{5}}{340} \approx 4.2 \times 10^{2}\ \text{N}$$

Key Takeaways

• With friction, $\text{GPE lost} = \text{KE gained} + \text{work done against friction}$.
• Average friction force can be found from dissipated energy divided by distance.

Common Mistakes

• Adding energies instead of subtracting: using $W_{\text{fr}} = \left|\Delta E_p\right| + \Delta E_k$.
• Forgetting that $\Delta E_p$ is negative and mishandling signs.
• Using vertical height instead of slope distance in $W = Fs$ (friction acts along the slope).

Things to Be Careful About

• The question asks for an average frictional force: you assume friction is approximately constant or use average work/average force.
• Make sure $s$ used in $F = W/s$ is the distance along AB (here $340\ \text{m}$).

Background Concept

For motion on an inclined plane, the component of weight down the slope is

$$mg\sin\theta$$

Resistive forces (friction + air resistance from a parachute) act opposite the direction of motion.

Newton’s second law along the slope is

$$\sum F = ma$$

where the sign of $a$ depends on your chosen positive direction.

Understanding the Question

From B to C:

• The slope angle is $20^{\circ}$ to the horizontal.
• Mass $m = 95\ \text{kg}$.
• The toboggan is still moving down the slope but is decelerating at a constant rate of $3.0\ \text{m s}^{-2}$.

You must find the frictional (resistive) force causing this deceleration.

Approach

1. Choose a positive direction (most naturally: down the slope).
2. Write the force balance along the slope: weight component minus resistive force equals $ma$.
3. Use $a=-3.0\ \text{m s}^{-2}$ because acceleration is up the slope when the motion is down the slope.
4. Solve for the resistive force.

Step-by-Step Reasoning

Let down-slope be positive.

Forces along the slope:

• Down slope: $mg\sin 20^{\circ}$.
• Up slope: resistive force $F$.

Newton’s second law:

$$mg\sin 20^{\circ} - F = ma$$

The toboggan is slowing down while moving down the slope, so acceleration is up the slope, hence $a=-3.0\ \text{m s}^{-2}$.

Rearrange:

$$F = mg\sin 20^{\circ} - ma = mg\sin 20^{\circ} + 3.0m$$

Now substitute values:

$$mg\sin 20^{\circ} = (95)(9.81)\sin 20^{\circ} \approx 931.95 \times 0.342 \approx 3.19 \times 10^{2}\ \text{N}$$ $$3.0m = 3.0 \times 95 = 2.85 \times 10^{2}\ \text{N}$$

So

$$F \approx 3.19 \times 10^{2} + 2.85 \times 10^{2} = 6.04 \times 10^{2}\ \text{N} \approx 6.0 \times 10^{2}\ \text{N}$$

Key Takeaways

• Deceleration means acceleration is opposite to velocity.
• On a slope, always resolve weight into components parallel and perpendicular to the plane.
• Use a consistent sign convention when applying $\sum F = ma$.

Common Mistakes

• Taking $a=+3.0\ \text{m s}^{-2}$ while also choosing down-slope as positive (sign error).
• Using $mg\cos 20^{\circ}$ instead of $mg\sin 20^{\circ}$ for the downslope component.
• Forgetting that the resistive force must be larger than $mg\sin 20^{\circ}$ to produce an up-slope acceleration.

Things to Be Careful About

• The frictional force here effectively includes parachute drag (a resistive force). The question still calls it “frictional force”, but it is the total resistive force along the slope.
• Quote the answer with a sensible number of significant figures (typically 2 s.f. to match $3.0$ and angles).