AP Physics 1 · Free-response skill

From a force-time graph to velocities and energy

The move. Read an impulse from a force-time graph, turn it into velocities with momentum conservation, and describe the collision in words and numbers.

How to do it

  1. Read the graph’s axes and units first. Impulse is the area under the force-time curve, so for a triangle use one-half times base times height, with the base in seconds.
  2. Turn impulse into motion with J=Δp=mΔvJ = \Delta p = m\Delta v for one object, then use conservation of momentum for the other, keeping signs for direction.
  3. Map each graph feature to a quantity: the slope of a velocity-time graph is the acceleration, and F=maF = ma links force to acceleration through the mass.
  4. When a question switches to the second object, ask which law links the forces the two objects exert on each other.
  5. To classify a collision, compute the total kinetic energy before and after and compare the two numbers in words.

Practice prompt Try it

On a level, frictionless air track, glider A (mass 0.40 kg) moves to the right at 0.75 m/s toward glider B (mass 0.25 kg), which is at rest. Bumpers on the gliders touch for 80 ms. A force sensor on glider B records the force that A exerts on B, shown in the graph. Take right as positive. No other horizontal forces act on the gliders.
Force exerted by glider A on glider B against time during the contact1020304050607080901000.511.522.533.540Time (ms)Force on B (N)
Force exerted by glider A on glider B against time during the contact A triangle. The force on B rises in a straight line from 0 N at t = 0 to a peak of 3.0 N at t = 40 ms, then falls in a straight line back to 0 N at t = 80 ms. The force points to the right the whole time.

Use the graph and the description to answer each part. Show your reasoning.

  1. Use the graph to find the impulse that glider A gives glider B during the contact.
  2. Find the velocity of glider B just after the contact.
  3. Find the velocity of glider A just after the contact.
  4. During the contact, at what time is glider B’s velocity changing most quickly? Give a reason based on the graph.
  5. Glider B pushes back on glider A during the contact. What is the largest size of that force, and which way does it point?
  6. Compute the total kinetic energy of the two gliders after the contact, and say whether the collision was elastic.

Write each part, then check yourself against the checklist. Nothing you type here is saved or graded.

Check yourself and see a model answer

Checklist

Our own checklist of what a full answer does, one line per point. Tick each one your response earns.

0 of 9 points

Model answer, annotated

(a) The impulse is the area of the triangle: J=12(0.080 s)(3.0 N)=0.12J = \frac{1}{2}(0.080 \text{ s})(3.0 \text{ N}) = 0.12 N·s, to the right.

  • Point 1 Uses the triangle area with the base converted from 80 ms to 0.080 s.

(b) B starts at rest, so 0.12=(0.25)vB0.12 = (0.25)v_B and vB=+0.48v_B = +0.48 m/s, to the right.

  • Point 2 Turns the impulse into a velocity with J=mΔvJ = m\Delta v.

(c) A gets an equal and opposite impulse, -0.12 N·s. Its momentum goes from (0.40)(0.75)=0.30(0.40)(0.75) = 0.30 kg·m/s to 0.30−0.12=0.180.30 - 0.12 = 0.18 kg·m/s, so vA=0.18/0.40=+0.45v_A = 0.18/0.40 = +0.45 m/s, still to the right.

  • Point 3 Uses momentum conservation (or the opposite impulse) with signs.

(d) B’s acceleration is the force on it divided by its mass, and the slope of its velocity-time graph is that acceleration. The force peaks at t = 40 ms, so B’s velocity changes most quickly at 40 ms.

  • Point 4 Gives the time of the peak, 40 ms.
  • Point 5 Links the steepest velocity change to the largest force on the graph.

(e) By Newton’s third law, B pushes on A with a force of the same size at every instant, so the largest force on A is 3.0 N. It points to the left, opposite to A’s motion.

  • Point 6 Gives the same peak size, 3.0 N.
  • Point 7 Gives the opposite direction, to the left.

(f) After: 12(0.40)(0.45)2+12(0.25)(0.48)2=0.0405+0.0288=0.0693\frac{1}{2}(0.40)(0.45)^2 + \frac{1}{2}(0.25)(0.48)^2 = 0.0405 + 0.0288 = 0.0693 J. Before, only A moved: 12(0.40)(0.75)2=0.1125\frac{1}{2}(0.40)(0.75)^2 = 0.1125 J. The total kinetic energy dropped, so the collision was not elastic.

  • Point 8 Adds both gliders’ kinetic energies after the contact.
  • Point 9 Compares the totals before and after to classify the collision.