Newton's Laws Practice Problems with Answers (AP Physics 1)

Practice problems

Work each one before opening the solution. Every step is shown, with the arithmetic left in so you can find where yours diverged.

1. 1. Net force and acceleration in one dimension

   A 17 kg sled is pushed along level ground by a horizontal 82 N force. Drag and friction together resist the motion with a combined 31 N acting backwards. Find the net force on the sled and its acceleration.

   Show the worked solution

   1. Take positive as the direction of the push. Only two forces act along that axis, so the vector sum is a signed sum.

      $$F_{\text{net}} = +82\ \text{N} - 31\ \text{N} = +51\ \text{N}$$

   2. The sled stays on the ground, so the vertical forces cancel and contribute nothing to the net force. The normal force and the weight are both real and both belong on the free-body diagram, but their sum along the vertical axis is zero.

   3. Now Newton's second law along the direction of travel.

      $$a = \frac{F_{\text{net}}}{m} = \frac{51\ \text{N}}{17\ \text{kg}} = 3.0\ \text{m/s}^2$$

   4. The sign is positive, so the acceleration points the same way as the push. That does not tell you the sled is moving forwards, only that whatever it is doing, its velocity is changing in the direction of the push.

   $F_{\text{net}} = 51\ \text{N}$ in the direction of the push, and $a = 3.0\ \text{m/s}^2$ in that same direction. The two vertical forces never entered the arithmetic, because they cancel.

2. 2. Equilibrium: a fixture on a single cable

   A 7.4 kg light fixture hangs from a single vertical cable attached to a ceiling. Find the tension in the cable (a) while the fixture hangs at rest, and (b) while a winch lowers the whole ceiling rig so that the fixture descends at a steady 0.30 m/s.

   Show the worked solution

   1. Take positive as up. Two forces act on the fixture: the cable's tension up and the gravitational force down. Its weight is

      $$mg = (7.4\ \text{kg})(9.8\ \text{m/s}^2) = 72.52\ \text{N}$$

   2. Part (a). At rest the fixture is not accelerating, so the net force is zero. That is translational equilibrium, essential knowledge 2.4.A.2.

      $$T - mg = 0 \quad \Rightarrow \quad T = 72.52\ \text{N}$$

   3. Part (b). Descending at a steady 0.30 m/s means the velocity is constant, so the acceleration is still zero. Zero acceleration gives zero net force, exactly as in part (a), and the algebra does not change at all.

      $$T - mg = m(0) \quad \Rightarrow \quad T = 72.52\ \text{N}$$

   4. Two significant figures: $73\ \text{N}$ in both cases. The speed 0.30 m/s appears nowhere in the working. It was given only to establish that the velocity was constant.

   5. The trap here is reading "moving down" as "accelerating down". Newton's first law separates the two: a nonzero velocity needs no force to sustain it, only a changing velocity does.

   $T = 73\ \text{N}$ in both parts. Moving at constant velocity and sitting still are the same problem, because both have zero acceleration and therefore zero net force. The tension only changes if the fixture speeds up or slows down, which is what problem 7 explores with a scale.

3. 3. Working backwards to a hidden force

   A 4.2 kg laboratory cart is pulled along a level bench by a horizontal 19 N force. Instead of the frictionless value you might expect, the cart is measured to accelerate at only 3.1 m/s^2 in the direction of the pull. Find the net force on the cart and the total resistive force acting on it.

   Show the worked solution

   1. Take positive as the direction of the pull. Newton's second law gives the net force straight from the measured acceleration, with no need to know what the individual forces are yet.

      $$F_{\text{net}} = ma = (4.2\ \text{kg})(3.1\ \text{m/s}^2) = 13.02\ \text{N}$$

   2. Now list what makes up that net force along the bench. There is the 19 N pull forwards and one unknown resistive force backwards.

      $$F_{\text{net}} = F_{\text{pull}} - F_{\text{resist}}$$

   3. Rearrange for the unknown and substitute.

      $$F_{\text{resist}} = F_{\text{pull}} - F_{\text{net}} = 19\ \text{N} - 13.02\ \text{N} = 5.98\ \text{N}$$

   4. Two significant figures: $6.0\ \text{N}$ backwards. As a check, a frictionless cart under the same pull would have accelerated at $19/4.2 = 4.5\ \text{m/s}^2$, and the measured 3.1 is lower, which is the direction the resistance should push the answer.

   $F_{\text{net}} = 13\ \text{N}$ forwards and the resistive force is $6.0\ \text{N}$ backwards. Newton's second law runs in both directions: forces give you an acceleration, and a measured acceleration gives you back the size of a force you never observed directly.

4. 4. Three forces in two dimensions

   Three horizontal forces act on a 6.0 kg crate sliding on frictionless ice, with directions given as seen from above: 38 N east, 14 N west, and 18 N north. Take east as positive $x$ and north as positive $y$. Find the magnitude and direction of the net force, then the crate's acceleration.

   Show the worked solution

   1. Newton's second law is a vector equation, so work one axis at a time. East and west lie on the $x$ axis; the 18 N force is entirely on the $y$ axis.

      $$\sum F_x = +38\ \text{N} - 14\ \text{N} = +24\ \text{N}$$
      $$\sum F_y = +18\ \text{N}$$

   2. Combine the two components with Pythagoras.

      $$\left|\vec{F}_{\text{net}}\right| = \sqrt{(24\ \text{N})^2 + (18\ \text{N})^2} = \sqrt{576 + 324} = \sqrt{900} = 30\ \text{N}$$

   3. The direction comes from the ratio of the components, measured from the positive $x$ axis.

      $$\theta = \arctan\!\left(\frac{18}{24}\right) = \arctan(0.75) = 36.87^\circ$$

      Both components are positive, so the net force points into the northeast quadrant: $36.9^\circ$ north of east.

   4. The acceleration is the net force divided by the mass, and it points along the net force because mass is a positive scalar.

      $$a = \frac{30\ \text{N}}{6.0\ \text{kg}} = 5.0\ \text{m/s}^2$$

   5. A common slip is to add the magnitudes, $38 + 14 + 18 = 70\ \text{N}$. That is more than twice the true answer. Forces add as vectors, so opposite directions subtract and perpendicular ones combine through a square root.

   $F_{\text{net}} = 30\ \text{N}$ at $37^\circ$ north of east, and $a = 5.0\ \text{m/s}^2$ in that same direction. The crate's velocity need not point that way: the net force sets the direction the velocity is changing, not the direction of travel.

5. 5. Tilted axes: a crate held on a frictionless ramp

   A 9.0 kg crate sits on a frictionless ramp inclined at $27^\circ$ above the horizontal. A rope runs from the crate straight up the slope, parallel to the ramp surface, and holds the crate at rest. Find the tension in the rope and the normal force from the ramp. Then find the crate's acceleration in the instant after the rope is cut.

   Show the worked solution

   1. Choose axes along and perpendicular to the ramp. Essential knowledge 2.2.B.4 says a coordinate system with one axis parallel to the direction of the acceleration simplifies the translation from free-body diagram to algebraic representation, and names an inclined plane as the example. Positive is down the slope, and perpendicular positive points away from the surface.

   2. The weight is the only force that needs resolving, since the tension is already along the slope and the normal force is already perpendicular to it.

      $$mg = (9.0\ \text{kg})(9.8\ \text{m/s}^2) = 88.2\ \text{N}$$
      $$mg\sin 27^\circ = (88.2\ \text{N})(0.45399) = 40.04\ \text{N} \quad \text{down the slope}$$
      $$mg\cos 27^\circ = (88.2\ \text{N})(0.89101) = 78.59\ \text{N} \quad \text{into the surface}$$

   3. Perpendicular to the ramp there is no acceleration, so those forces balance and the normal force matches the perpendicular component of the weight.

      $$F_N = mg\cos 27^\circ = 78.59\ \text{N}$$

      It is not 88.2 N. On a ramp the surface only has to support the perpendicular part.

   4. Along the slope the crate is at rest, so the net force is zero. The tension acts up the slope, which is the negative direction here.

      $$mg\sin 27^\circ - T = 0 \quad \Rightarrow \quad T = 40.04\ \text{N}$$

   5. Cut the rope and the tension term vanishes. Nothing else changed, so the along-slope equation becomes

      $$a = \frac{mg\sin 27^\circ}{m} = g\sin 27^\circ = (9.8\ \text{m/s}^2)(0.45399) = 4.449\ \text{m/s}^2$$

      The mass cancels, so every object on a frictionless $27^\circ$ ramp accelerates at the same rate.

   6. Two significant figures throughout: $T = 40\ \text{N}$, $F_N = 79\ \text{N}$, $a = 4.4\ \text{m/s}^2$ down the slope.

   $T = 40\ \text{N}$ up the slope, $F_N = 79\ \text{N}$, and after the cut $a = 4.4\ \text{m/s}^2$ down the slope. The normal force never equalled the 88 N weight, and it does not change when the rope is cut, because the rope was parallel to the surface and had no perpendicular component to remove.

6. 6. Equilibrium in two dimensions: a sign on two cables

   A 12.0 kg sign hangs from two cables anchored to a ceiling. The two cables are symmetric about the sign's centre and each makes an angle of $35.0^\circ$ with the horizontal. Find the tension in each cable, and compare it with half the sign's weight.

   Show the worked solution

   1. Take positive $x$ as horizontal and positive $y$ as up. The sign is at rest, so the net force is zero on both axes at once. By symmetry the two tensions have the same magnitude $T$.

      $$mg = (12.0\ \text{kg})(9.8\ \text{m/s}^2) = 117.6\ \text{N}$$

   2. Horizontal axis first, because it settles the symmetry claim. One cable pulls with $T\cos 35.0^\circ$ to the left and the other with $T\cos 35.0^\circ$ to the right. They cancel for any value of $T$, so the horizontal axis gives no new information and confirms the two tensions must be equal.

   3. Vertical axis. Each cable contributes an upward component $T\sin 35.0^\circ$, and together they must carry the whole weight.

      $$2T\sin 35.0^\circ - mg = 0$$

   4. Solve for the tension.

      $$T = \frac{mg}{2\sin 35.0^\circ} = \frac{117.6\ \text{N}}{2(0.57358)} = \frac{117.6\ \text{N}}{1.14715} = 102.5\ \text{N}$$

   5. Compare that with half the weight, $58.8\ \text{N}$. The tension is about 1.74 times larger, because each cable only devotes $\sin 35.0^\circ$ of itself to holding the sign up. The rest, $T\cos 35.0^\circ = 84.0\ \text{N}$ per cable, goes into a horizontal tug of war that the two cables fight to a draw.

   6. Check the vertical sum: $2(102.5\ \text{N})(0.57358) = 117.6\ \text{N}$, which is the weight exactly.

   $T = 103\ \text{N}$ in each cable, against a half weight of only 58.8 N. Flatten the cables towards horizontal and the tension climbs without limit, since $\sin\theta$ heads for zero in the denominator. That is why a washing line pulled taut can snap under a light load, and why truly horizontal cables cannot support anything at all.

7. 7. A lift, and what the scale actually reads

   A 58.0 kg student stands on a bathroom scale in a lift. Take up as positive. Find the scale reading when the lift (a) accelerates upward at 1.40 m/s^2, (b) travels downward at a constant 3.0 m/s, and (c) accelerates downward at 2.60 m/s^2. (d) What downward acceleration would make the scale read zero?

   Show the worked solution

   1. A scale reads the force it pushes up on you with, which is the normal force. The CED puts it plainly: the magnitude of the apparent weight of a system is the magnitude of the normal force exerted on the system (essential knowledge 2.6.C.1). Your gravitational weight never changes in this problem.

      $$mg = (58.0\ \text{kg})(9.8\ \text{m/s}^2) = 568.4\ \text{N}$$

   2. Set up Newton's second law once, with up positive, and reuse it for every part.

      $$F_N - mg = ma \quad \Rightarrow \quad F_N = m(g + a)$$

   3. Part (a), accelerating up, so $a = +1.40\ \text{m/s}^2$.

      $$F_N = (58.0\ \text{kg})(9.8 + 1.40)\ \text{m/s}^2 = (58.0)(11.20) = 649.6\ \text{N}$$

   4. Part (b), constant velocity downwards, so $a = 0$. The direction of travel is irrelevant; only the acceleration enters the equation.

      $$F_N = (58.0\ \text{kg})(9.8 + 0)\ \text{m/s}^2 = 568.4\ \text{N}$$

   5. Part (c), accelerating down, so $a = -2.60\ \text{m/s}^2$. The sign convention set at the start does the work here.

      $$F_N = (58.0\ \text{kg})(9.8 - 2.60)\ \text{m/s}^2 = (58.0)(7.20) = 417.6\ \text{N}$$

   6. Part (d). Setting $F_N = 0$ gives $0 = m(g + a)$, so $a = -g = -9.8\ \text{m/s}^2$: the lift is in free fall. The student is then weightless in the CED's sense, 2.6.C.3, because gravity is the only force acting. The gravitational force on the student is still 568 N throughout.

   7. Three significant figures: $650\ \text{N}$, $568\ \text{N}$, $418\ \text{N}$, and free fall for the zero reading.

   (a) $650\ \text{N}$, (b) $568\ \text{N}$, (c) $418\ \text{N}$, (d) a downward acceleration of $9.8\ \text{m/s}^2$, which is free fall. The student's mass and the gravitational force on her were identical in all four parts. Only the normal force moved, and the normal force is what a scale reports (2.6.C.1 and 2.6.C.2).

8. 8. A push angled downwards, where the normal force exceeds mg

   A 14 kg crate is pushed along a level floor by a 60 N force directed $25^\circ$ below the horizontal, as if you were leaning into it. The coefficient of kinetic friction between crate and floor is $\mu_k = 0.19$. Find the normal force, the friction force, and the crate's acceleration.

   Show the worked solution

   1. Take positive $x$ as the direction of travel and positive $y$ as up. Split the applied force into components. Pushing below the horizontal means the vertical component points down.

      $$F_x = (60\ \text{N})\cos 25^\circ = (60)(0.90631) = 54.38\ \text{N}$$
      $$F_y = (60\ \text{N})\sin 25^\circ = (60)(0.42262) = 25.36\ \text{N} \quad \text{downwards}$$

   2. The crate stays on the floor, so the vertical forces balance. Three of them act now: the normal force up, the weight down, and the push's downward component.

      $$F_N - mg - F_y = 0$$

   3. The weight is $mg = (14\ \text{kg})(9.8\ \text{m/s}^2) = 137.2\ \text{N}$, so

      $$F_N = mg + F_y = 137.2\ \text{N} + 25.36\ \text{N} = 162.6\ \text{N}$$

      You are pressing the crate into the floor, so the floor pushes back harder. This is the mirror image of an angled pull, which reduces the normal force instead.

   4. Friction responds to that larger normal force.

      $$F_{f,k} = \mu_k F_N = (0.19)(162.6\ \text{N}) = 30.89\ \text{N}$$

   5. Newton's second law along the floor, using only the horizontal component of the push.

      $$a = \frac{F_x - F_{f,k}}{m} = \frac{54.38\ \text{N} - 30.89\ \text{N}}{14\ \text{kg}} = \frac{23.49\ \text{N}}{14\ \text{kg}} = 1.678\ \text{m/s}^2$$

   6. Two significant figures: $1.7\ \text{m/s}^2$. Had you written $F_N = mg$ out of habit, friction would have come out as $(0.19)(137.2) = 26.07\ \text{N}$ and the acceleration as $2.0\ \text{m/s}^2$, about 21 percent too high. Every number after the normal force inherits that error.

   $F_N = 163\ \text{N}$, $F_{f,k} = 31\ \text{N}$, and $a = 1.7\ \text{m/s}^2$ in the direction of the push. The normal force sits 25 N above the crate's 137 N weight, because the push has a downward component. Steepen the push and you gain vertical load faster than you gain horizontal drive, until at some angle the crate stops accelerating altogether.

9. 9. Two connected blocks on a frictionless table

   A 5.0 kg block and a 3.0 kg block sit on a frictionless horizontal table, joined by a light inextensible string. A 36 N horizontal force pulls on the 5.0 kg block, dragging the 3.0 kg block along behind it. (a) Find the acceleration of the pair and the tension in the string. (b) Repeat for the same 36 N force applied to the 3.0 kg block instead, with the 5.0 kg block trailing.

   Show the worked solution

   1. Take positive as the direction of the pull. Start with the two blocks as one system. The string tension is internal to that system, so it does not appear, and the only external horizontal force is the 36 N pull.

      $$a = \frac{F}{m_1 + m_2} = \frac{36\ \text{N}}{5.0\ \text{kg} + 3.0\ \text{kg}} = \frac{36\ \text{N}}{8.0\ \text{kg}} = 4.5\ \text{m/s}^2$$

   2. Part (a), tension. To see an internal force you must break the system apart, so take the trailing 3.0 kg block on its own. The string is the only horizontal force on it.

      $$T = m_{\text{trailing}}\, a = (3.0\ \text{kg})(4.5\ \text{m/s}^2) = 13.5\ \text{N}$$

   3. Check on the other block: the 5.0 kg block feels the 36 N pull forwards and the string pulling 13.5 N backwards, so its net force is $36 - 13.5 = 22.5\ \text{N}$, and $22.5/5.0 = 4.5\ \text{m/s}^2$. The two blocks agree, as they must, because the string does not stretch.

   4. Part (b). The system is unchanged, so the acceleration is unchanged at $4.5\ \text{m/s}^2$. Only the tension moves, because now the string has to accelerate the 5.0 kg block.

      $$T = (5.0\ \text{kg})(4.5\ \text{m/s}^2) = 22.5\ \text{N}$$

   5. So the tension is set entirely by the mass on the far side of the string. It is never half the applied force, and it is never the applied force either.

   (a) $a = 4.5\ \text{m/s}^2$ and $T = 14\ \text{N}$. (b) The same $a = 4.5\ \text{m/s}^2$, but $T = 23\ \text{N}$. The string only has to accelerate what is behind it, so pulling the light block puts nearly twice the tension in the same string. Whichever way round you pull, the tension is smaller than the 36 N applied force.

10. 10. An Atwood machine

    Two blocks hang from the ends of a light string that passes over an ideal pulley: 6.2 kg on one side and 3.8 kg on the other. The system is released from rest. Find the magnitude of the acceleration and the tension in the string.

    Show the worked solution

    1. The ideal string and ideal pulley are what make this tractable. Essential knowledge 2.3.A.3.ii says the tension in an ideal string is the same at all points, and 2.3.A.3.iv defines an ideal pulley as one with negligible mass turning with negligible friction. So one symbol $T$ describes both sides, and both blocks share one acceleration magnitude.

    2. Take positive as the direction the system actually moves: down for the 6.2 kg block and up for the 3.8 kg block. With that choice both blocks have the same signed acceleration $+a$.

    3. Heavy block, positive down:

       $$m_1 g - T = m_1 a$$

       Light block, positive up:

       $$T - m_2 g = m_2 a$$

    4. Add the two equations. The tension cancels, which is the whole reason for adding them.

       $$m_1 g - m_2 g = (m_1 + m_2)a$$
       $$a = \frac{(m_1 - m_2)g}{m_1 + m_2} = \frac{(6.2 - 3.8)(9.8)}{6.2 + 3.8} = \frac{(2.4)(9.8)}{10.0} = \frac{23.52}{10.0} = 2.352\ \text{m/s}^2$$

    5. Put that back into either equation. Using the light block:

       $$T = m_2(g + a) = (3.8\ \text{kg})(9.8 + 2.352) = (3.8)(12.152) = 46.18\ \text{N}$$

    6. Confirm with the heavy block, which is an independent route to the same number:

       $$T = m_1(g - a) = (6.2\ \text{kg})(9.8 - 2.352) = (6.2)(7.448) = 46.18\ \text{N}$$

    7. Two significant figures: $a = 2.4\ \text{m/s}^2$ and $T = 46\ \text{N}$. Sanity check: the two weights are $m_2 g = 37.24\ \text{N}$ and $m_1 g = 60.76\ \text{N}$, and 46 N sits between them. It has to. A tension above both weights would accelerate both blocks upwards, and a tension below both would drop them both.

    $a = 2.4\ \text{m/s}^2$, with the 6.2 kg block descending, and $T = 46\ \text{N}$ throughout the string. Notice that $2T = 92\ \text{N}$ is less than the 98 N total weight of the two blocks: the pulley holds up less than the system weighs precisely because the system is accelerating.

11. 11. Modified Atwood with friction on the table

    A 4.5 kg block sits on a rough horizontal table and is joined by a light string over an ideal pulley at the table's edge to a 2.8 kg block hanging freely. The coefficients between the table block and the table are $\mu_s = 0.31$ and $\mu_k = 0.24$. Show that the system moves when released from rest, then find the acceleration and the string tension.

    Show the worked solution

    1. First check whether anything happens at all. The table block is on a level surface with nothing pressing on it vertically except gravity and the table, so

       $$F_N = m_1 g = (4.5\ \text{kg})(9.8\ \text{m/s}^2) = 44.1\ \text{N}$$
       $$F_{f,s,\text{max}} = \mu_s F_N = (0.31)(44.1\ \text{N}) = 13.67\ \text{N}$$

    2. The force trying to drag the system into motion is the hanging block's weight, $m_2 g = (2.8)(9.8) = 27.44\ \text{N}$. That is well above the 13.67 N ceiling static friction can supply, so the system slides and every friction force from here is kinetic.

    3. Take positive as the direction of motion: the hanging block moves down and the table block moves towards the pulley. Kinetic friction opposes the table block's motion, so it points backwards, away from the pulley.

       $$F_{f,k} = \mu_k F_N = (0.24)(44.1\ \text{N}) = 10.584\ \text{N}$$

    4. Write Newton's second law for each block.

       Hanging block: $\quad m_2 g - T = m_2 a$

       Table block: $\quad T - F_{f,k} = m_1 a$

    5. Add them so the tension cancels.

       $$m_2 g - F_{f,k} = (m_1 + m_2)a$$
       $$a = \frac{27.44\ \text{N} - 10.584\ \text{N}}{4.5\ \text{kg} + 2.8\ \text{kg}} = \frac{16.856\ \text{N}}{7.3\ \text{kg}} = 2.309\ \text{m/s}^2$$

    6. Back-substitute into the hanging block's equation.

       $$T = m_2(g - a) = (2.8\ \text{kg})(9.8 - 2.309) = (2.8)(7.491) = 20.97\ \text{N}$$

    7. Check on the table block, which uses different numbers entirely:

       $$T = F_{f,k} + m_1 a = 10.584\ \text{N} + (4.5\ \text{kg})(2.309\ \text{m/s}^2) = 10.584 + 10.391 = 20.97\ \text{N}$$

    8. Two significant figures: $a = 2.3\ \text{m/s}^2$ and $T = 21\ \text{N}$.

    The system moves, because the hanging block's 27.4 N weight beats the 13.7 N that static friction can supply. Then $a = 2.3\ \text{m/s}^2$ and $T = 21\ \text{N}$. The tension is well below the hanging block's 27.4 N weight, which is the tell that the block is accelerating downwards rather than hanging in equilibrium.

12. 12. Explain why: which force is bigger in a truck and car collision

    A loaded delivery truck of mass 8200 kg collides head on with a 1300 kg car. The car is wrecked and its driver is thrown violently forwards, while the truck's cab is barely marked. A student concludes that the truck must have exerted a larger force on the car than the car exerted on the truck. Say whether that conclusion is correct, and explain the damage. Then, taking the magnitude of the force between the two vehicles at one instant to be $3.4 \times 10^5\ \text{N}$, find each vehicle's acceleration at that instant.

    Show the worked solution

    1. The conclusion is wrong, and the CED says so in one line. Essential knowledge 2.3.A.1 gives Newton's third law as

       $$\vec{F}_{A \text{ on } B} = -\vec{F}_{B \text{ on } A}$$

       The two forces in the pair are equal in magnitude and opposite in direction at every instant of the collision, no matter the masses, the speeds, or which vehicle was moving. There is no version of this where one of them is larger.

    2. So why the difference in damage? Because damage tracks acceleration, and the pair members act on different objects with different masses. Take the direction the truck was travelling as positive, so the force on the truck from the car is negative and the force on the car from the truck is positive.

    3. Truck:

       $$a_{\text{truck}} = \frac{-3.4 \times 10^5\ \text{N}}{8200\ \text{kg}} = -41.46\ \text{m/s}^2$$

    4. Car:

       $$a_{\text{car}} = \frac{+3.4 \times 10^5\ \text{N}}{1300\ \text{kg}} = +261.5\ \text{m/s}^2$$

    5. The ratio of the two magnitudes is exactly the inverse ratio of the masses:

       $$\frac{261.5}{41.46} = 6.31 = \frac{8200}{1300}$$

       Same force, six times the acceleration, because the car has one sixth of the mass. Occupants are injured by their own acceleration, so the car's occupants take roughly six times the punishment from an interaction that treated the two vehicles identically.

    6. One last check that catches the standard confusion. The two forces in a third-law pair never cancel, because they act on different objects. Only forces on the same object appear together on one free-body diagram, and only those can cancel. The truck's diagram shows one arrow from this interaction; the car's diagram shows the other.

    The conclusion is wrong: the two forces are equal in magnitude and opposite in direction at every instant, which is essential knowledge 2.3.A.1. The accelerations are what differ. At the stated instant the truck accelerates at $41\ \text{m/s}^2$ and the car at $2.6 \times 10^2\ \text{m/s}^2$, a factor of 6.3 apart, matching the inverse ratio of their masses. Equal forces, unequal masses, unequal accelerations, and that is what the wreckage records.