Dawson College · 203-SN1-RE

Dawson 203-SN1-RE: mechanics tutoring

Work with an online tutor on motion, forces, energy and momentum in Dawson 203-SN1-RE. Start with a question or concept you want explained. Your tutor can help you draw the situation, identify the forces and choose the physical law to use. If you have a diagram or calculation, you can review it together.

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Course overview

Dawson 203-SN1-RE begins the physics sequence with motion, forces, and conservation laws.

Your 203-SN1-RE revision route

Choose only topics listed for your assessment. The course description supports this subject, but does not establish your section’s exam date or chapter cutoff.

Choose topics. Review displacement, velocity and acceleration
Follow a worked example. Separate projectile motion into components: worked examples
Try without solutions. Choose the existing forces or momentum diagnostic if covered
Get help with a step. Ask about 203-SN1-RE tutoring Include the chapter, assessment date and question; attempted work is useful but not required. Sessions are currently in English.

Official course source Your section outline and instructor announcements take priority. These are Nablio study routes and exercises, not official exams.

Practise the reasoning between steps

Choose a workshop when the topic is included in your instructor’s announced material. These original exercises are not an exam reproduction or a complete course review.

Connect equilibrium and static friction

When rigid-body equilibrium is in your assigned material, practise force diagrams, torques and the friction limit in one model. This supplement is not a complete course review or an exam prediction.

Ladder equilibrium: maximum distance and minimum angle

Projectile workshop: one clock, two components

This original practice develops two-dimensional mechanics. Use it if projectile motion is on your course outline. It is not a prediction of your section’s midterm questions or coverage. Keep the existing collision-and-spring example for a separate conservation-law review.

Model each projectile as a particle near Earth’s surface. Neglect air resistance, take constant $g=9.8\,\mathrm{m/s^2}$ , and consider motion only from release until the specified first contact. Set $t=0$ at release, positive $x$ to the right and positive $y$ upward. Distances are in metres and times in seconds. Record the launch and landing heights in the same coordinate system.

Build the model before choosing a root

x=x_0+v_{0x}t,\qquad y=y_0+v_{0y}t-\tfrac12gt^2.

v_x=v_{0x},\qquad v_y=v_{0y}-gt.

For a launch speed $v_0$ at angle $\theta$ above the positive horizontal, $v_{0x}=v_0\cos\theta$ and $v_{0y}=v_0\sin\theta$ . Horizontal velocity stays constant; vertical velocity changes. Both components use the same elapsed time. At the highest point, only $v_y=0$ ; the projectile need not be at rest.

The vertical component is drawn at the tip of the horizontal component to form a vector triangle. It does not represent a second stage of the flight.

Worked example 1: horizontal launch from a ledge

A ball leaves a ledge horizontally at $12\,\mathrm{m/s}$ , from $19.6\,\mathrm m$ above level ground. Find its flight time, horizontal displacement, and velocity just before impact. Predict the sign of its vertical impact velocity first.

Show the horizontal-launch solution

Set $x_0=0$ , $y_0=19.6$ , $v_{0x}=12$ , $v_{0y}=0$ , and ground height $y=0$ . Vertical motion determines the landing time:

0=19.6-4.9t^2\quad\Longrightarrow\quad t=\pm2\,\mathrm s.

Use $t=2\,\mathrm s$ . The negative root is before release and lies outside our model interval. Horizontal motion then gives $\Delta x=12(2)=24\,\mathrm m$ .

\mathbf v_{\rm impact}=(12\,\hat{\mathbf i}-19.6\,\hat{\mathbf j})\,\mathrm{m/s}.

The speed is $\sqrt{12^2+19.6^2}\approx22.98\,\mathrm{m/s}$ . Its direction is $\arctan(19.6/12)\approx58.52^\circ$ below the positive horizontal. Report the reference direction, not just an unsigned angle. The vertical velocity is negative, as predicted.

Check independently: $v^2=v_0^2+2g(19.6)$ . This confirms the speed, but does not determine the direction by itself. A horizontal launch has zero initial vertical velocity, not zero vertical acceleration.

Worked example 2: an upward launch lands lower

A projectile is released from $y_0=24.5\,\mathrm m$ with components $v_{0x}=14.7\,\mathrm{m/s}$ and $v_{0y}=19.6\,\mathrm{m/s}$ . The ground is at $y=0$ , with no obstacle before impact. These components correspond to speed $24.5\,\mathrm{m/s}$ and launch angle $\arctan(4/3)$ . Find the landing time, horizontal displacement and impact direction.

Show the unequal-height solution

0=24.5+19.6t-4.9t^2\quad\Longrightarrow\quad t^2-4t-5=(t-5)(t+1)=0.

The roots are $5\,\mathrm s$ and $-1\,\mathrm s$ . Only $5\,\mathrm s$ belongs to the post-release flight. Thus $\Delta x=14.7(5)=73.5\,\mathrm m$ .

v_x=14.7,\qquad v_y=19.6-9.8(5)=-29.4\quad(\mathrm{m/s}).

The impact speed is $\sqrt{14.7^2+29.4^2}\approx32.87\,\mathrm{m/s}$ , at $\arctan(29.4/14.7)\approx63.43^\circ$ below the positive horizontal.

The apex occurs at $t=19.6/9.8=2\,\mathrm s$ , at height $44.1\,\mathrm m$ . The projectile returns to launch height at $4\,\mathrm s$ , then continues to the lower ground until $5\,\mathrm s$ . Therefore $2v_{0y}/g$ is not the total flight time here. Likewise, the equal-height range formula is inapplicable. The energy check gives $v^2=24.5^2+2(9.8)(24.5)=1080.45\,\mathrm{m^2/s^2}$ .

Four graduated transfer problems

Use the same ideal model and $g=9.8\,\mathrm{m/s^2}$ . For every problem, write the target height equation before calculating. Keep roots until you have checked the event they describe.

1. Separate launch speed from fall time

A ball leaves a $4.9\,\mathrm m$ -high ledge horizontally at $6\,\mathrm{m/s}$ , landing on level ground. Find its flight time and horizontal displacement. If the launch speed doubles, which answer changes?

Hint for problem 1

There is no horizontal velocity in the vertical landing equation.

Solution to problem 1

$0=4.9-4.9t^2$ gives the physical time $t=1\,\mathrm s$ . Then $\Delta x=6\,\mathrm m$ . Doubling the horizontal launch speed leaves the fall time at $1\,\mathrm s$ and doubles the displacement to $12\,\mathrm m$ . This conclusion uses the no-drag model and the unchanged launch height.

2. Two positive roots do not mean two landings

A projectile starts at $(0,0)$ with $v_{0x}=10\,\mathrm{m/s}$ , $v_{0y}=14.7\,\mathrm{m/s}$ . A thin horizontal platform at height $9.8\,\mathrm m$ begins at $x=15\,\mathrm m$ and extends to the right. There is no obstacle for $x<15\,\mathrm m$ . At which time and position does the projectile first contact the platform?

Hint for problem 2

Find both times at the platform’s height, then check the horizontal position and vertical velocity at each one.

Solution to problem 2

$9.8=14.7t-4.9t^2$ gives $(t-1)(t-2)=0$ . At $t=1\,\mathrm s$ , the projectile is at $x=10\,\mathrm m$ , rising with $v_y=4.9\,\mathrm{m/s}$ , where no platform exists. At $t=2\,\mathrm s$ , it is at $x=20\,\mathrm m$ , descending with $v_y=-4.9\,\mathrm{m/s}$ . First contact is therefore $(20,9.8)\,\mathrm m$ at $2\,\mathrm s$ . At the platform’s leading edge, $t=1.5\,\mathrm s$ and $y=11.025\,\mathrm m>9.8\,\mathrm m$ , so it clears the edge. Both time roots are physically meaningful height crossings; geometry identifies the landing.

3. Reject a root for a reason

A projectile starts at height $9.8\,\mathrm m$ , with $v_{0x}=8\,\mathrm{m/s}$ and $v_{0y}=4.9\,\mathrm{m/s}$ . Find the ground-impact time and displacement. Explain why one algebraic root is excluded.

Hint for problem 3

Divide the landing equation by $4.9$ . The coordinate value $y=0$ is not the launch position.

Solution to problem 3

$0=9.8+4.9t-4.9t^2$ gives $(t-2)(t+1)=0$ . The root $-1\,\mathrm s$ lies before the specified release at $t=0$ ; use $t=2\,\mathrm s$ . Thus $\Delta x=16\,\mathrm m$ . As a check, $v_y=4.9-9.8(2)=-14.7\,\mathrm{m/s}$ , consistent with downward impact.

4. Reconstruct an impact velocity

A projectile travels $30\,\mathrm m$ to the right and lands $9.8\,\mathrm m$ below its launch point after $2\,\mathrm s$ . Find its initial velocity components, then the speed and direction just before impact.

Hint for problem 4

Use displacement equations to recover the initial components. The vertical displacement divided by time is an average velocity, not the impact velocity.

Solution to problem 4

$v_{0x}=30/2=15\,\mathrm{m/s}$ . Vertically, $-9.8=2v_{0y}-4.9(2)^2$ , so $v_{0y}=4.9\,\mathrm{m/s}$ . At impact, $v_x=15$ and $v_y=4.9-9.8(2)=-14.7$ , in $\mathrm{m/s}$ . The speed is $\sqrt{15^2+14.7^2}\approx21.00\,\mathrm{m/s}$ , at $\arctan(14.7/15)\approx44.42^\circ$ below the positive horizontal. The average vertical velocity is $-4.9\,\mathrm{m/s}$ ; using it as the final component would be an error.

Choose your next attempt from the first error

Wrong component or sign: redraw the launch triangle and axes before retrying problem 1.
Wrong root: list time, horizontal position and vertical direction for each candidate in problem 2.
Wrong flight-time formula: compare the launch and landing heights in worked example 2.
Wrong impact angle: draw the final velocity vector, then state the angle’s reference direction as in problem 4.

Bring your course outline, test date, component sketch and first height equation to tutoring. Ask for help with the first decision you cannot explain, then solve a fresh problem without the worked solution. Tutoring is offered in English; this French/English material does not imply French-language tutoring.

203-SN1-RE topics to review

KinematicsNewton’s lawsWork and energyMomentumRotation and torque

Common difficulties

Choosing forces, momentum or energy from the physical interaction rather than a familiar formula
Recognizing that momentum conservation in a collision does not imply kinetic-energy conservation

Mechanics: choose forces or momentum

Find the acceleration of a sliding block, then analyse a sticking collision. Choose the physical principle before substituting numbers.

Before you start: Allow 20–25 minutes to try both questions, then 15 minutes to check and retry any missed steps.

1. Net force on a sliding block

A $4.0\,\mathrm{kg}$ block is already sliding right on a horizontal surface. A horizontal $18\,\mathrm N$ pull acts right. The coefficient of kinetic friction is $\mu_k=0.20$ . Use $g=9.8\,\mathrm{m/s^2}$ . Find the acceleration.

Show the explained solution

Vertically, $N=mg=39.2\,\mathrm N$ . Kinetic friction is leftward because the block slides right.
$f_k=\mu_kN=7.84\,\mathrm N.$
With right positive, $ma=18-7.84$ , so $a=2.54\,\mathrm{m/s^2}$ to the right. The pull itself is not the net force.

Force diagrams and friction

2. Two carts stick together

A $0.60\,\mathrm{kg}$ cart moving right at $3.0\,\mathrm{m/s}$ hits and sticks to a stationary $0.90\,\mathrm{kg}$ cart. Neglect external horizontal impulse during the collision. Find their shared velocity and the kinetic energy lost.

Show the explained solution

For both carts together, horizontal momentum is conserved: $(0.60)(3.0)=(1.50)v_f.$
Thus $v_f=1.20\,\mathrm{m/s}$ to the right. With $K=\tfrac12mv^2$ , $K_i=\tfrac12(0.60)(3.0)^2=2.70\,\mathrm J,\qquad K_f=\tfrac12(1.50)(1.20)^2=1.08\,\mathrm J.$
The loss is $1.62\,\mathrm J$ . Momentum conservation does not imply kinetic-energy conservation in a sticking collision.

Review momentum and energy in a sticking collision

Use your answers to plan your revision

For a force error
Draw only forces acting on the selected body, choose axes and write the net-force equation before inserting values.
For a collision error
Name the system and collision interval. State the external-impulse assumption before conserving momentum.

You can ask for help with either question above, even if you are unsure how to begin. Your tutor can help you draw the forces, choose a system and explain when to use Newton’s laws or conservation of momentum.

Request 203-SN1-RE exam preparation help

Original Nablio practice, not an official examination.

Course topic sources

Dawson Physics: Mechanics 203-SN1-RE

Preparing for this course

Review vector components, trigonometry, equation rearrangement and SI units. Read position, velocity and acceleration graphs by distinguishing height, slope and area. Before solving a mechanics problem, identify the system and time interval; conservation laws can apply during one stage of an event and fail during another.

Registration requirements

The guidance above is study preparation, not an eligibility decision. Confirm current prerequisites, restrictions and availability with Dawson College.

Check the official course catalogue

Try a course problem

On a horizontal frictionless track, a cart of mass $m_1=1\,\mathrm{kg}$ , initially moving at $u_1=6\,\mathrm{m/s}$ , strikes and sticks to a stationary cart of mass $m_2=2\,\mathrm{kg}$ , whose initial velocity is $u_2=0$ . After the brief collision, the pair reaches an initially uncompressed ideal spring of stiffness $k=300\,\mathrm{N/m}$ , fixed at its far end to a wall. Find the joined carts’ speed just after impact, their maximum spring compression and the kinetic energy lost during impact. Neglect external horizontal impulse during the collision.

Three stages on a horizontal frictionless track. Before impact, cart m₁ = 1 kg moves right at u₁ = 6 m/s toward stationary cart m₂ = 2 kg, with u₂ = 0. Both are separated from an uncompressed ideal spring of stiffness k = 300 N/m attached to a wall. Just after impact the carts are joined, move right with unknown common speed v, and still do not touch the spring. At maximum compression they are momentarily at rest against the shortened spring. A dashed vertical reference marks the uncompressed spring’s free end; x measures from that reference to the compressed spring’s free end. The wall stays fixed. The drawing is not to scale. — Read the stages in time order. The collision happens before spring contact. Here v is the common speed just after impact. The dashed reference marks the spring’s uncompressed free end; x measures spring compression, not the carts’ travel since impact. Arrows above the carts show velocity, not force; their lengths are not to scale.

Show the reasoning

1. Treat the collision first. Choose the incoming direction as positive and let $v$ be the common velocity just after impact. For the two-cart system, negligible external horizontal impulse means total momentum is conserved: $m_1u_1+m_2u_2=(m_1+m_2)v$ . Substitution gives $(1)(6)+(2)(0)=3v$ , so $v=2\,\mathrm{m/s}$ . 2. Account for the impact energy loss. Let $K_i$ and $K_f$ be the total kinetic energies immediately before and after the collision. Then $K_i=\frac12m_1u_1^2+\frac12m_2u_2^2=18\,\mathrm{J}$ , and $K_f=\frac12(m_1+m_2)v^2=\frac12(3)(2^2)=6\,\mathrm{J}$ . The kinetic energy lost is $K_i-K_f=12\,\mathrm{J}$ , converted to internal energy and other forms during sticking. Momentum conservation does not imply kinetic-energy conservation in this inelastic collision. 3. Start a separate spring-compression stage. Let $x\ge0$ be the maximum compression. At that instant the carts momentarily stop, so their post-impact kinetic energy has become spring potential energy. With no friction and an ideal spring, $\frac12(m_1+m_2)v^2=\frac12kx^2$ . Therefore $x^2=(m_1+m_2)v^2/k=(3)(2^2)/300=0.04\,\mathrm{m^2}$ , giving $x=0.20\,\mathrm{m}$ . 4. Check which energy is available. The spring stores $\frac12(300)(0.20)^2=6\,\mathrm{J}$ , matching the energy after the collision. Using the original $18\,\mathrm{J}$ would ignore the impact loss.

Your turn: double the incoming speed

Use the same frictionless track, sticking collision and ideal spring, but change the incoming speed to $u_1=12\,\mathrm{m/s}$ . Keep $m_1=1\,\mathrm{kg}$ , $m_2=2\,\mathrm{kg}$ , $u_2=0$ and $k=300\,\mathrm{N/m}$ . Neglect external horizontal impulse during the brief collision. Find the common speed just after impact, the maximum spring compression and the kinetic energy lost during impact. Before calculating, predict which answers double and which quadruple. Name the conservation law for each stage, with the answer closed.

Check your reasoning

1. Use momentum during impact
Let $v$ denote the common velocity, positive in the incoming direction. Momentum conservation gives $(1)(12)+(2)(0)=3v$ , so $v=4\,\mathrm{m/s}$ . The speed doubles.
2. Compare the kinetic energies
Let $K_i$ and $K_f$ be the kinetic energies immediately before and after impact. They are $K_i=\tfrac12(1)(12)^2=72\,\mathrm{J}$ and $K_f=\tfrac12(3)(4)^2=24\,\mathrm{J}$ . The impact loss is $K_i-K_f=48\,\mathrm{J}$ , four times the original loss. The two kinetic energies also quadruple.
3. Start a new energy stage
Only the post-impact $24\,\mathrm{J}$ is available for spring compression. With $x\ge0$ denoting maximum compression, $\tfrac12(300)x^2=24$ gives $x=0.40\,\mathrm{m}$ , twice the original compression. Check: the spring energy is $150(0.40)^2=24\,\mathrm{J}$ . If you used 72 J here, you carried energy lost during impact into the next stage. The track removes frictional loss, but it does not make the sticking collision elastic.

Angular momentum: choose the system, axis and interval

For angular-momentum problems in your current chapter, name the system and rotation axis before checking the external angular impulse. Work through a sticking collision, then check kinetic energy separately.

How tutoring can help

Your tutor can help you draw a force diagram or check one you have started for missing forces. Together, you can decide whether Newton’s laws, energy or momentum applies to each stage of a problem. For rotation, your tutor can explain how the chosen axis determines the lever arms and torque signs, then check your torque equation against the diagram.

Exam preparation for 203-SN1-RE

Use your section’s outline to choose kinematics, Newton’s-law, energy, momentum and rotation questions for the next exam; check laboratory requirements there too. Split multi-stage problems into time intervals and state which physical principle applies in each. Check the units of your answer and whether its size and direction make sense for the situation.

We work through permitted practice and comparable problems, without completing graded assignments or exams on your behalf.

Independent practice

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