AP Physics C: Mechanics Newton's Laws of Motion — Worked Answer Explanations

Unit 2 · 12 questions explained

1. Question 1 · Easy

   A 5 kg block is pushed across a frictionless surface by a net force $F=15$ N. What is the acceleration?

   • A
     $2$ m/s²

     Why not A: Divided force by an incorrect mass of 7.5 kg.
   • B
     $3$ m/s²Correct
   • C
     $5$ m/s²

     Why not C: Divided $F$ by $g$ instead of $m$.
   • D
     $75$ m/s²

     Why not D: Multiplied $F$ by $m$ instead of dividing.
   Explanation

   Newton's second law: $a=\frac{F_{net}}{m}=\frac{15}{5}=3$ m/s².

   Key takeaway

   $F_{net} = ma$ — always use net force and total mass.

2. Question 2 · Easy

   A 2 kg block hangs from a rope attached to the ceiling. What is the tension in the rope? ($g=10$ m/s²)

   • A
     $10$ N

     Why not A: Used only $g$ without multiplying by mass.
   • B
     $20$ NCorrect
   • C
     $2$ N

     Why not C: Set tension equal to mass in kilograms.
   • D
     $0$ N

     Why not D: Assumed the rope exerts no force because the block is in equilibrium.
   Explanation

   The block is in equilibrium: $T-mg=0\Rightarrow T=mg=2\times 10=20$ N. By Newton's third law, the rope pulls up with 20 N and the block pulls down on the rope with 20 N.

   Key takeaway

   For a hanging mass in equilibrium, tension equals weight. Always draw a free-body diagram first.

3. Question 3 · Easy

   Two blocks, $m_1=3$ kg and $m_2=5$ kg, are connected by a massless rope and pulled across a frictionless surface by a force $F=16$ N applied to $m_2$. What is the tension in the rope between them?

   • A
     $6$ NCorrect
   • B
     $10$ N

     Why not B: Applied $F=ma$ to $m_1$ alone using the total acceleration but set $m=5$ kg.
   • C
     $2$ m/s² (wrong unit)

     Why not C: Reported the acceleration instead of tension.
   • D
     $16$ N

     Why not D: Set tension equal to applied force, ignoring the acceleration of the system.
   Explanation

   System acceleration: $a=\frac{F}{m_1+m_2}=\frac{16}{8}=2$ m/s². Tension pulls $m_1$: $T=m_{1}a=3\times 2=6$ N.

   Key takeaway

   Find system acceleration first, then isolate one block to solve for the internal rope tension.

4. Question 4 · Easy

   A block of mass $m$ sits on a plane inclined at angle $\theta$ (frictionless). Which expression gives the acceleration along the incline?

   • A
     $g\cos \theta$

     Why not A: Used the normal-force component instead of the along-incline component.
   • B
     $g\sin \theta$Correct
   • C
     $g\tan \theta$

     Why not C: Divided the two components rather than identifying the relevant one.
   • D
     $g$

     Why not D: Ignored the geometric projection onto the incline.
   Explanation

   Along the incline: $ma=mg\sin \theta$, so $a=g\sin \theta$. The normal force $N=mg\cos \theta$ does no work along the incline. The mass cancels, so acceleration is independent of $m$.

   Key takeaway

   On a frictionless incline, resolve gravity along the slope: $a = g\sin\theta$, independent of mass.

5. Question 5 · Medium

   A particle of mass $m$ experiences a velocity-dependent drag force $F_{drag}=-bv$, where $b>0$. Starting from rest, which expression gives $v(t)$?

   • A
     $v(t)=\frac{mg}{b}\left(1-e^{-bt/m}\right)$Correct
   • B
     $v(t)=\frac{mg}{b}{e}^{-bt/m}$

     Why not B: Missing the $(1-\cdot )$ factor; this form doesn't satisfy $v(0)=0$.
   • C
     $v(t)=\frac{mg}{b}\left(1+e^{-bt/m}\right)$

     Why not C: Sign error in the exponential term gives $v(0)=2mg/b\ne 0$.
   • D
     $v(t)=\frac{mg}{b}\left(e^{bt/m}-1\right)$

     Why not D: Exponential has positive exponent, so $v\to \infty$ rather than terminal velocity.
   Explanation

   Newton's second law: $m\frac{dv}{dt}=mg-bv$. Separate variables:
   $${\int }_0^v\frac{d{v}^{\prime }}{mg-b{v}^{\prime }}={\int }_0^t\frac{d{t}^{\prime }}{m}.$$
   $$-\frac{1}{b}\ln \frac{mg-bv}{mg}=\frac{t}{m}\Rightarrow v(t)=\frac{mg}{b}\left(1-e^{-bt/m}\right).$$
   As $t\to \infty$, $v\to v_T=mg/b$ (terminal velocity).

   Key takeaway

   Linear drag leads to exponential approach to terminal velocity. Solve the separable ODE $m\,dv/dt = mg - bv$ and apply $v(0) = 0$.

6. Question 6 · Medium

   A block of mass $m$ on a surface with kinetic friction coefficient ${\mu }_k$ is pushed by a horizontal force $F$. What is the net acceleration?

   • A
     $\frac{F}{m}$

     Why not A: Ignored friction.
   • B
     $\frac{F-{\mu }_{k}mg}{m}$Correct
   • C
     $\frac{F}{m}-{\mu }_k{g}^2$

     Why not C: Dimensional error: subtracted ${\mu }_k{g}^2$ (wrong units).
   • D
     $\frac{F+{\mu }_{k}mg}{m}$

     Why not D: Added friction force instead of subtracting it (wrong direction).
   Explanation

   Normal force $N=mg$ (horizontal surface). Friction: $f_k={\mu }_{k}mg$ opposing motion. Net force: $F-{\mu }_{k}mg=ma$. So $a=\frac{F-{\mu }_{k}mg}{m}=\frac{F}{m}-{\mu }_{k}g$.

   Key takeaway

   Draw the FBD: normal force balances weight on a horizontal surface, so $f_k = \mu_k mg$. Subtract it from $F$ before dividing by $m$.

7. Question 7 · Medium

   An Atwood machine has masses $m_1=4$ kg and $m_2=6$ kg connected by a massless rope over a frictionless pulley. What is the acceleration of the system? ($g=10$ m/s²)

   • A
     $1$ m/s²

     Why not A: Divided $g$ by the total mass $10$ kg.
   • B
     $2$ m/s²Correct
   • C
     $5$ m/s²

     Why not C: Used only the mass difference divided by $g$ (dimensional error).
   • D
     $4$ m/s²

     Why not D: Used only $m_1$ in the denominator.
   Explanation

   For the Atwood machine: $a=\frac{(m_2-m_1)g}{m_1+m_2}=\frac{(6-4)(10)}{4+6}=\frac{20}{10}=2$ m/s². The heavier mass descends.

   Key takeaway

   Atwood machine formula: $a = \dfrac{\Delta m}{m_{total}}g$. Both masses appear in the denominator because both are accelerated.

8. Question 8 · Medium

   A car rounds a banked curve of radius $R$ at angle $\theta$ (no friction). What is the speed $v$ at which no friction is needed?

   • A
     $\sqrt{Rg\sin \theta }$

     Why not A: Used $\sin \theta$ instead of $\tan \theta$.
   • B
     $\sqrt{Rg\cos \theta }$

     Why not B: Used $\cos \theta$ instead of $\tan \theta$.
   • C
     $\sqrt{Rg\tan \theta }$Correct
   • D
     $Rg\tan \theta$

     Why not D: Forgot the square root.
   Explanation

   Normal force components: $N\cos \theta =mg$ (vertical) and $N\sin \theta =m{v}^2/R$ (centripetal). Dividing: $\tan \theta =v^2/(Rg)\Rightarrow v=\sqrt{Rg\tan \theta }$.

   Key takeaway

   Divide vertical and horizontal FBD equations for a banked curve to eliminate $N$ and isolate $v^2 = Rg\tan\theta$.

9. Question 9 · Hard

   A particle of mass $m$ falls under gravity through a medium with quadratic drag $F_{drag}=-c{v}^2$. What is the terminal velocity $v_T$?

   • A
     $\sqrt{\frac{mg}{c}}$Correct
   • B
     $\frac{mg}{c}$

     Why not B: Terminal velocity for linear drag, not quadratic.
   • C
     $\sqrt{\frac{c}{mg}}$

     Why not C: Inverted the ratio inside the square root.
   • D
     $\frac{c}{mg}$

     Why not D: Inverted and omitted the square root.
   Explanation

   At terminal velocity $a=0$: $mg-c{v}_T^2=0\Rightarrow v_T^2=\frac{mg}{c}\Rightarrow v_T=\sqrt{\frac{mg}{c}}$.

   Note: $v_T\propto \sqrt{m}$ for quadratic drag vs. $v_T\propto m$ for linear drag — an important distinction for Physics C.

   Key takeaway

   Terminal velocity occurs when net force $= 0$. Set drag equal to weight and solve: quadratic drag gives $v_T = \sqrt{mg/c}$.

10. Question 10 · Hard

    A block of mass $m$ is pushed against a vertical wall by a horizontal force $F$ (with $F>{\mu }_{s}mg$). What is the minimum $F$ needed to prevent sliding?

    • A
      $F_{min}=mg$

      Why not A: Ignored friction; the block can't be supported by $F$ alone with no friction.
    • B
      $F_{min}={\mu }_{s}mg$

      Why not B: Set friction force equal to ${\mu }_{s}mg$ without recognizing that $N=F$.
    • C
      $F_{min}=\frac{mg}{{\mu }_s}$Correct
    • D
      $F_{min}=\frac{{\mu }_s}{mg}$

      Why not D: Inverted the ratio and didn't check units.
    Explanation

    The horizontal $F$ provides the normal force on the wall: $N=F$. Maximum static friction (upward): $f_s={\mu }_{s}N={\mu }_{s}F$. For equilibrium vertically: ${\mu }_{s}F\ge mg\Rightarrow F_{min}=\frac{mg}{{\mu }_s}$.

    Key takeaway

    When a block is pressed against a vertical wall, $N = F_{applied}$ (horizontal), so $f_s = \mu_s F$. Set $f_s = mg$ to find $F_{min}$.

11. Question 11 · Hard

    A particle of mass $m$ on a frictionless horizontal surface is attached to a spring (constant $k$) and experiences a drag force $F_d=-bv$. Which differential equation governs its motion?

    • A
      $m\ddot{x}+kx=0$

      Why not A: Omits the damping term $b\dot{x}$.
    • B
      $m\ddot{x}-b\dot{x}+kx=0$

      Why not B: Sign error: drag opposes motion, so $F_d=-bv$ contributes $+b\dot{x}$ not $-b\dot{x}$ when moved to the left side.
    • C
      $m\ddot{x}+b\dot{x}+kx=0$Correct
    • D
      $m\ddot{x}+b\dot{x}-kx=0$

      Why not D: Spring force sign error: spring pulls back toward equilibrium, giving $-kx$.
    Explanation

    Net force: $F_{net}=-kx-bv=-kx-b\dot{x}$. Newton's second law: $m\ddot{x}=-kx-b\dot{x}$. Rearranging: $m\ddot{x}+b\dot{x}+kx=0$. This is the standard damped harmonic oscillator equation. When $b^2<4mk$, the system is underdamped and oscillates with decreasing amplitude.

    Key takeaway

    The damped oscillator ODE is $m\ddot{x} + b\dot{x} + kx = 0$. The signs come directly from Newton's 2nd law: both restoring and drag forces act against displacement/velocity.

12. Question 12 · Hard

    A rocket of initial mass $M_0$ expels exhaust at speed $u$ relative to the rocket. Applying Newton's second law to variable mass, what is the thrust force?

    • A
      $F_{thrust}=u\frac{dM}{dt}$ (positive)

      Why not A: $dM/dt<0$ (mass decreasing), so this expression is negative — wrong sign for thrust direction.
    • B
      $F_{thrust}=-u\frac{dM}{dt}$Correct
    • C
      $F_{thrust}=u{M}_0$

      Why not C: Uses initial mass as a constant rather than the rate of mass change.
    • D
      $F_{thrust}=\frac{u}{dM/dt}$

      Why not D: Divided rather than multiplied $u$ by the mass flow rate.
    Explanation

    From the momentum impulse equation for variable-mass systems (Tsiolkovsky), the thrust is:
    $$F_{thrust}=-u\frac{dM}{dt}.$$
    Since exhaust is expelled ($dM/dt<0$), $F_{thrust}>0$, pointing in the direction of motion. The rocket equation is then $M\frac{dv}{dt}=-u\frac{dM}{dt}$, which integrates to $\Delta v=u\ln \frac{M_0}{M_f}$.

    Key takeaway

    Rocket thrust $= -u(dM/dt)$. Since $dM/dt < 0$ (mass leaving), thrust is positive. The Tsiolkovsky equation $\Delta v = u\ln(M_0/M_f)$ follows by separating variables.