Collège Ahuntsic · 203-SN1-RE

Tutoring for Ahuntsic 203-SN1-RE Mechanics

Get help connecting forces, acceleration and motion. Try the short example below, or ask a tutor to help you understand the question and choose a first step.

Your Ahuntsic mechanics course

Ahuntsic 203-SN1-RE Mechanics connects motion, forces and conservation laws. You use differential calculus to describe change, build models of physical situations and check your results using units, measurements and physical reasoning.

This guide focuses on one force-and-motion connection. Use your current course outline to choose the rest of your revision topics.

Course topics

  • Translational and rotational motion
  • Newton’s laws and translational and rotational dynamics
  • Conservation of energy, momentum and angular momentum

Start with the object and the interval

Choose the object you are studying and a positive direction. Mark each time when the force changes. For each interval, find the acceleration from the net force and the mass before choosing a motion equation.

Two distinctions to practise

  • A moving object can have zero net force and zero acceleration. Its velocity then stays constant.
  • When a force changes, carry the velocity from the end of one interval to the beginning of the next.

What happens when the push ends?

A 2 kg cart moves right at 1 m/s along a horizontal track. Neglect friction and air resistance, and model the cart as a particle. A constant horizontal force of 4 N pushes it right for 2 s, then is removed. Find the cart’s velocity at 5 s and its total displacement over those 5 s. Take right as positive. First predict whether the cart stops when the push ends.

Show a hint

Split the motion at 2 s. Calculate the velocity at that instant, then use it as the initial velocity of the second interval. That second interval lasts 5−2=35-2=3 s.

Show the worked example
  1. 1. Identify the forces in each interval

    The cart is the system. Its weight acts downward and the track’s normal force acts upward. There is no vertical acceleration, so these forces balance. During the push, the net horizontal force is 4 N; after the push, it is zero.

  2. 2. Find the velocity when the push ends

    For the first interval, let v0v_0 be the initial velocity, v1v_1 the final velocity, a1a_1 the acceleration and Δt1\Delta t_1 the elapsed time. The acceleration is constant and equals the net force divided by the mass: a1=4 N2 kg=2 m/s2.a_1=\frac{4\,\mathrm{N}}{2\,\mathrm{kg}}=2\,\mathrm{m/s^2}. Hence v1=v0+a1Δt1=1 m/s+(2 m/s2)(2 s)=5 m/s.\begin{aligned}v_1&=v_0+a_1\Delta t_1\\&=1\,\mathrm{m/s}\\&\quad+(2\,\mathrm{m/s^2})(2\,\mathrm{s})\\&=5\,\mathrm{m/s}.\end{aligned} The positive result means the cart is moving right.

  3. 3. Carry that velocity into the second interval

    After 2 s, the net horizontal force is zero. The acceleration in this second interval, a2a_2, is therefore zero. The cart continues at 5 m/s for the next 3 s. Its velocity at 5 s is therefore +5 m/s+5\,\mathrm{m/s}. The acceleration changes when the push ends, but the velocity does not change instantly.

  4. 4. Add the two displacements

    For one interval, let Δx\Delta x be the displacement, viv_i the velocity at the start of the interval, Δt\Delta t its duration and aa the constant acceleration during that interval. Use Δx=viΔt+12a(Δt)2\Delta x=v_i\Delta t+\tfrac12a(\Delta t)^2 separately on each constant-acceleration interval. Let Δx1\Delta x_1 and Δx2\Delta x_2 be the displacements in the first and second intervals. With SI values, Δx1=1(2)+12(2)(2)2=6 m,Δx2=5(3)=15 m.\begin{aligned}\Delta x_1&=1(2)+\tfrac12(2)(2)^2\\&=6\,\mathrm{m},\\\Delta x_2&=5(3)=15\,\mathrm{m}.\end{aligned} The total displacement is 6+15=216+15=21 m to the right. Both intervals have positive velocity, so the distance travelled is also 21 m.

  5. 5. Check the result with work and energy

    Let WW be the work done by the push and ΔK\Delta K the change in kinetic energy. Using SI values, W=4(6)=24W=4(6)=24 J. The kinetic-energy calculation gives ΔK=12(2)(52−12)=24 J.\Delta K=\tfrac12(2)(5^2-1^2)=24\,\mathrm{J}. The values agree, as required by the work–energy theorem. In the second interval, the net work is zero and the kinetic energy stays constant. In SI units, N m=J\mathrm{N\,m}=\mathrm{J}.

Your turn: change the second interval

Keep the first 2 s unchanged. Instead of removing the push and leaving zero net force, suppose a constant net horizontal force of 2 N now acts left for the next 3 s. Find the velocity at 5 s and the total displacement from 0 to 5 s. Does the cart reverse direction during this interval?

Show a hint

Keep right positive: the new net force is −2-2 N. The second interval still starts at +5 m/s+5\,\mathrm{m/s}. Find when the velocity would first reach zero before deciding whether the cart turns around.

Check the reasoning
  1. 1. Calculate the new acceleration and velocity

    The second-interval acceleration is a2=−2 N2 kg=−1 m/s2.a_2=\frac{-2\,\mathrm{N}}{2\,\mathrm{kg}}=-1\,\mathrm{m/s^2}. Let vv be the velocity at time t=5t=5 s, after 3 s under the new force: v=5 m/s+(−1 m/s2)(3 s)=2 m/s.\begin{aligned}v&=5\,\mathrm{m/s}\\&\quad+(-1\,\mathrm{m/s^2})(3\,\mathrm{s})\\&=2\,\mathrm{m/s}.\end{aligned} The velocity is still positive: the cart is moving right while slowing down.

  2. 2. Calculate displacement over the correct interval

    Δx2=5(3)+12(−1)(3)2=10.5 m.\begin{aligned}\Delta x_2&=5(3)+\tfrac12(-1)(3)^2\\&=10.5\,\mathrm{m}.\end{aligned} Add the unchanged first displacement: 6+10.5=16.56+10.5=16.5 m to the right.

  3. 3. Check whether a reversal occurs

    If the same acceleration continued, let Δtstop\Delta t_{\mathrm{stop}} be the time from the force change to rest. Then Δtstop=0−5−1=5 s.\Delta t_{\mathrm{stop}}=\frac{0-5}{-1}=5\,\mathrm{s}. Only 3 s have elapsed since that change, so the cart has not stopped or reversed. A negative acceleration does not by itself mean leftward motion.

If you predicted an immediate stop, revisit the connection between net force and acceleration. If you used 5 s in the second interval, mark the start and end times before substituting.

Work through forces and motion with a tutor

Bring a question, even if you do not know how to begin. Your tutor can help you choose the object to study, draw its forces, choose axes and write the first equation. If you have already started, you can review your diagram and equations together, then try the same idea on a new problem.

A tutor can check the direction and units with you, connect each symbol to the quantity it represents, and help you explain whether the result makes physical sense.

Build a useful practice set

Choose one force problem, one motion-graph question and one work–energy question from the topics your class has covered. Before calculating, name the system, positive direction and time interval. After checking a solution, redo the setup without looking at it.

Angular momentum: choose the system and axis

Your mechanics course includes angular-momentum conservation. Start with the system and axis, then check the external angular impulse before choosing a conservation equation. The six questions move from that choice to changing inertia and a lump sticking to a disk.

Try six angular-momentum questions

Bring a question to a tutor. You can build the diagram and first angular-momentum equation together, or review what you have started. You can check the external-torque assumption, the signs and the energy comparison before retrying a missed question.

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