Newton's Second Law (College Board AP® Physics 1: Algebra-Based): Revision Note

A rocket of mass $0.8\times {10}^5\mathrm{kg}$ produces an upward thrust of $15\mathrm{MN}$ and has a weight of $8\mathrm{MN}$. When in flight, the force due to air resistance is $500\mathrm{kN}$.

What is acceleration of the rocket?

A      $-187.5\mathrm{m}/{\mathrm{s}}^2$

B      $-81\mathrm{m}/{\mathrm{s}}^2$

C     $81\mathrm{m}/{\mathrm{s}}^2$

D      $87.5\mathrm{m}/{\mathrm{s}}^2$

The correct answer is C

Answer:

Step 1: Analyze the scenario

• The direction of motion is upwards, therefore upwards is the positive direction

  • Air resistance (downward acting), ${\overrightarrow{F}}_f=-500\mathrm{kN}=-500\times {10}^3\mathrm{N}$

  • Weight (downward acting), ${\overrightarrow{F}}_g=-8\mathrm{MN}=-8\times {10}^6\mathrm{N}$

  • Thrust (upward acting), ${\overrightarrow{F}}_{thrust}=15\mathrm{MN}=15\times {10}^6\mathrm{N}$

Step 2: Calculate the net external force

${\overrightarrow{F}}_{net}=\left(15\times {10}^6\right)+\left(-8\times {10}^6\right)+\left(-500\times {10}^3\right)$

${\overrightarrow{F}}_{net}=6.5\times {10}^6\mathrm{N}=6.5\mathrm{MN}$

• The positive value indicates that the resultant force acts in the direction of motion, i.e. upwards

Step 3: State the equation for Newton's second law

${\overrightarrow{a}}_{sys}=\frac{{\overrightarrow{F}}_{net}}{m_{sys}}$

Step 4: Calculate the acceleration and state the direction

$\overrightarrow{a}=\frac{6.5\times {10}^6}{0.8\times {10}^5}$

$\overrightarrow{a}=81\mathrm{m}/{\mathrm{s}}^2$ upwards

• The answer is therefore C

A car windscreen makes an angle of $30°$ to the horizontal. The weight of a water drop accelerating down the windscreen is $8.0\times {10}^{-5}\mathrm{N}$. The normal contact force on the water drop is $F_N$.

Which of the following shows the correct magnitude of the acceleration of the water drop, assuming that the friction between the windscreen and the drop is negligible?

A      $0.05\mathrm{m}/{\mathrm{s}}^2$

B      $0.8\mathrm{m}/{\mathrm{s}}^2$

C      $1\mathrm{m}/{\mathrm{s}}^2$

D      $5\mathrm{m}/{\mathrm{s}}^2$

The correct answer is D

Answer:

Step 1: Analyze the scenario

• The water drop is accelerating downward, parallel to the line of the slope

  • So there is a net external force acting down parallel to the line of the slope

• The water drop has no motion perpendicular to the line of the slope

  • So there is a zero net external force perpendicular to the line of the slope

• Therefore, only the motion of the water drop parallel to the line of the slope should be considered

Step 2: Draw a diagram to show the component acting parallel to the inclined plane

• The water drop moves down the windscreen as a result of its weight component parallel to the slope of the windscreen

• Draw a diagram to show the correct component

Step 3: Determine the net external force of the water drop moving down the windscreen

${\overrightarrow{F}}_{net}={\overrightarrow{F}}_g\sin \left(30\right)$

${\overrightarrow{F}}_{net}=\left(8.0\times {10}^{-5}\right)\left(\sin \left(30\right)\right)$

${\overrightarrow{F}}_{net}=4.0\times {10}^{-5}\mathrm{N}$

Step 4: Determine the mass of the water drop

• Recall the equation for calculating the weight of an object

${\overrightarrow{F}}_g=mg$

• Rearrange the equation to make mass, $m$ the subject:

$m=\frac{{\overrightarrow{F}}_g}{g}$

• Calculate the mass of the water drop:

$m=\frac{\left(8.0\times {10}^{-5}\right)}{10}$

$m=8.0\times {10}^{-6}\mathrm{kg}$

Step 5: Calculate the acceleration of the water drop moving down the windscreen

• Recall the equation for Newton's second law of motion:

${\overrightarrow{a}}_{sys}=\frac{{\overrightarrow{F}}_{net}}{m_{sys}}$

• Calculate the acceleration of the water drop:

${\overrightarrow{a}}_{sys}=\frac{\left(4.0\times {10}^{-5}\right)}{\left(8.0\times {10}^{-6}\right)}$

${\overrightarrow{a}}_{sys}=5\mathrm{m}/{\mathrm{s}}^2$

• The answer is therefore D