Hard Further Maths Further dynamics Questions

Challenging, exam-style Further Maths Further dynamics questions with worked solutions. Stretch yourself on the hardest further-dynamics, variable-force, a=v dv/dx, resistance problems.

further-dynamicsvariable-forcea=v dv/dxresistancea=dv/dtterminal-speed
Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
A van of mass 24002400 kg moves along a straight horizontal road. The engine of the van works at a constant rate of 2800028000 W. When the van is moving with speed vv m s−1^{-1} the total resistance to motion has magnitude 7v27v^{2} N. Find the maximum speed of the van. Give your answer to 3 significant figures. Select the correct value from the options given.
Show worked solution

Worked solution

  1. Relate the driving force to the power and the speed

    F=PvF=\dfrac{P}{v}

    The engine works at a constant rate, so the driving force is P/vP/v.

  2. Apply the condition for the maximum speed

    a=0 ⇒ Pv=7v2a=0\ \Rightarrow\ \dfrac{P}{v}=7v^{2}

    At the maximum speed the driving force just balances the resistance.

  3. Rearrange for the speed

    v3=P7=280007v^{3}=\dfrac{P}{7}=\dfrac{28000}{7}

    Multiplying up gives a power of vv equal to P/kP/k.

  4. Recall the two forms of the acceleration

    a=dvdt=vdvdxa=\dfrac{dv}{dt}=v\dfrac{dv}{dx}

    The form to use depends on whether the force is given in terms of time or of displacement.

  5. Recall Newton's second law

    F=maF=ma

    The resultant force equals the mass times the acceleration at every instant.

  6. Recall the work done by a variable force

    W=∫F dxW=\int F\,dx

    The work of a force that varies with position is the integral of the force with respect to displacement.

  7. Recall the work-energy principle

    W=12mv2−12mu2W=\tfrac{1}{2}mv^{2}-\tfrac{1}{2}mu^{2}

    The work done by the resultant force equals the change in kinetic energy.

  8. Recall the impulse of a variable force

    J=∫F dtJ=\int F\,dt

    The impulse of a time-varying force is the integral of the force with respect to time.

  9. Recall the impulse-momentum principle

    J=mv−muJ=mv-mu

    The impulse of the resultant force equals the change in momentum.

  10. Recall that power is the rate of working

    P=FvP=Fv

    For a driving force FF at speed vv the power developed is FvFv.

  11. Recall the condition for the maximum speed

    a=0a=0

    At the maximum speed the acceleration is zero, so the driving force balances the resistance.

  12. Separate the variables before integrating

    ∫dvf(v)=∫dx\int \frac{dv}{f(v)}=\int dx

    A separable first-order equation is solved by collecting the speed on one side.

  13. State the value of gg used throughout

    g=9.8 m s−2g=9.8\ \text{m s}^{-2}

    All numerical answers use this value of the acceleration due to gravity.

  14. Check the units of each term

    [F]=N,[W]=J[F]=\text{N},\quad [W]=\text{J}

    Forces are in newtons and work and energy in joules.

  15. Note that a resistance opposes the motion

    R=−kvnR=-kv^{n}

    A resistive force always acts opposite to the velocity, so it reduces the speed.

  16. Solve for the maximum speed

    v=2800073=15.9 m s−1v=\sqrt[3]{\dfrac{28000}{7}}=15.9\ \text{m s}^{-1}

    Taking the (3)(3)th root gives the maximum speed.

Answer
v=15.9 m s−1v=15.9\ \text{m s}^{-1}
Question 2
9 markschallenging
A small disc of mass 66 kg moves in a straight line. The only force acting on the small disc is a resistance of magnitude 2v22v^{2} N, where vv m s−1^{-1} is the speed of the small disc. The small disc passes a point OO with speed 3030 m s−1^{-1}. Using a=dvdta=\dfrac{dv}{dt}, find the time taken for the speed of the small disc to decrease to 55 m s−1^{-1}. Select the correct value from the options given.
Show worked solution

Worked solution

  1. Choose the form of the acceleration for a time question

    a=dvdta=\dfrac{dv}{dt}

    We want a time, so use the acceleration as dv/dtdv/dt.

  2. Write down the equation of motion

    mdvdt=−2v2m\dfrac{dv}{dt}=-2v^{2}

    The resistance is the only force and it opposes the motion.

  3. Separate the variables and integrate

    ∫mv2 dv=−∫2 dt\int\dfrac{m}{v^{2}}\,dv=-\int 2\,dt

    The equation is separable in vv and tt.

  4. Carry out the integration

    −mv=−2 t+C-\dfrac{m}{v}=-2\,t+C

    The integral of 1/v21/v^{2} is −1/v-1/v.

  5. Recall the two forms of the acceleration

    a=dvdt=vdvdxa=\dfrac{dv}{dt}=v\dfrac{dv}{dx}

    The form to use depends on whether the force is given in terms of time or of displacement.

  6. Recall Newton's second law

    F=maF=ma

    The resultant force equals the mass times the acceleration at every instant.

  7. Recall the work done by a variable force

    W=∫F dxW=\int F\,dx

    The work of a force that varies with position is the integral of the force with respect to displacement.

  8. Recall the work-energy principle

    W=12mv2−12mu2W=\tfrac{1}{2}mv^{2}-\tfrac{1}{2}mu^{2}

    The work done by the resultant force equals the change in kinetic energy.

  9. Recall the impulse of a variable force

    J=∫F dtJ=\int F\,dt

    The impulse of a time-varying force is the integral of the force with respect to time.

  10. Recall the impulse-momentum principle

    J=mv−muJ=mv-mu

    The impulse of the resultant force equals the change in momentum.

  11. Recall that power is the rate of working

    P=FvP=Fv

    For a driving force FF at speed vv the power developed is FvFv.

  12. Recall the condition for the maximum speed

    a=0a=0

    At the maximum speed the acceleration is zero, so the driving force balances the resistance.

  13. Separate the variables before integrating

    ∫dvf(v)=∫dx\int \frac{dv}{f(v)}=\int dx

    A separable first-order equation is solved by collecting the speed on one side.

  14. State the value of gg used throughout

    g=9.8 m s−2g=9.8\ \text{m s}^{-2}

    All numerical answers use this value of the acceleration due to gravity.

  15. Check the units of each term

    [F]=N,[W]=J[F]=\text{N},\quad [W]=\text{J}

    Forces are in newtons and work and energy in joules.

  16. Note that a resistance opposes the motion

    R=−kvnR=-kv^{n}

    A resistive force always acts opposite to the velocity, so it reduces the speed.

  17. Recall that momentum is mass times velocity

    p=mvp=mv

    An impulse changes the momentum of the particle.

  18. Apply the limits and solve for the time

    t=62(15−130)=0.5 st=\dfrac{6}{2}\left(\dfrac{1}{5}-\dfrac{1}{30}\right)=0.5\ \text{s}

    Evaluating between the two speeds gives the time.

Answer
t=0.5 st=0.5\ \text{s}
Question 3
9 markschallenging
A ball bearing of mass 88 kg moves along a straight horizontal line. The only force acting on the ball bearing is a resistance of magnitude 3v23v^{2} N, where vv m s−1^{-1} is the speed of the ball bearing. The ball bearing passes a point OO with speed 3636 m s−1^{-1}. Using a=vdvdxa=v\dfrac{dv}{dx}, find the distance travelled by the ball bearing as its speed decreases to 99 m s−1^{-1}. Give your answer to 3 significant figures. Select the correct value from the options given.
Show worked solution

Worked solution

  1. Choose the form of the acceleration for a speed-dependent force

    a=vdvdxa=v\dfrac{dv}{dx}

    The resistance depends on the speed and we want distance, so use v dv/dxv\,dv/dx.

  2. Write down the equation of motion

    mvdvdx=−3v2m v\dfrac{dv}{dx}=-3v^{2}

    The resistance is the only force and it opposes the motion, hence the minus sign.

  3. Cancel a factor of vv and separate the variables

    mv dv=−3 dx\dfrac{m}{v}\,dv=-3\,dx

    Dividing by v2v^{2} and multiplying by vv leaves dv/vdv/v on the left.

  4. Integrate both sides

    mln⁡v=−3 x+Cm\ln v=-3\,x+C

    The integral of 1/v1/v is ln⁡v\ln v.

  5. Recall the two forms of the acceleration

    a=dvdt=vdvdxa=\dfrac{dv}{dt}=v\dfrac{dv}{dx}

    The form to use depends on whether the force is given in terms of time or of displacement.

  6. Recall Newton's second law

    F=maF=ma

    The resultant force equals the mass times the acceleration at every instant.

  7. Recall the work done by a variable force

    W=∫F dxW=\int F\,dx

    The work of a force that varies with position is the integral of the force with respect to displacement.

  8. Recall the work-energy principle

    W=12mv2−12mu2W=\tfrac{1}{2}mv^{2}-\tfrac{1}{2}mu^{2}

    The work done by the resultant force equals the change in kinetic energy.

  9. Recall the impulse of a variable force

    J=∫F dtJ=\int F\,dt

    The impulse of a time-varying force is the integral of the force with respect to time.

  10. Recall the impulse-momentum principle

    J=mv−muJ=mv-mu

    The impulse of the resultant force equals the change in momentum.

  11. Recall that power is the rate of working

    P=FvP=Fv

    For a driving force FF at speed vv the power developed is FvFv.

  12. Recall the condition for the maximum speed

    a=0a=0

    At the maximum speed the acceleration is zero, so the driving force balances the resistance.

  13. Separate the variables before integrating

    ∫dvf(v)=∫dx\int \frac{dv}{f(v)}=\int dx

    A separable first-order equation is solved by collecting the speed on one side.

  14. State the value of gg used throughout

    g=9.8 m s−2g=9.8\ \text{m s}^{-2}

    All numerical answers use this value of the acceleration due to gravity.

  15. Check the units of each term

    [F]=N,[W]=J[F]=\text{N},\quad [W]=\text{J}

    Forces are in newtons and work and energy in joules.

  16. Note that a resistance opposes the motion

    R=−kvnR=-kv^{n}

    A resistive force always acts opposite to the velocity, so it reduces the speed.

  17. Apply the limits and solve for the distance

    x=83ln⁡ ⁣(369)=3.70 mx=\dfrac{8}{3}\ln\!\left(\dfrac{36}{9}\right)=3.70\ \text{m}

    Evaluating between the two speeds gives the distance.

Answer
x=3.70 mx=3.70\ \text{m}
Question 4
9 markschallenging
A particle PP of mass 33 kg moves along the positive xx-axis. When the particle PP is at a distance xx metres from the origin OO, it is acted on by a resultant force of magnitude F=(40−3x2)F=(40-3x^{2}) N directed in the direction of increasing xx. The particle PP passes through OO moving in the direction of increasing xx with speed 33 m s−1^{-1}. Using a=vdvdxa=v\dfrac{dv}{dx}, find the speed of the particle PP when x=3x=3. Give your answer to 3 significant figures. Select the correct value from the options given.
Show worked solution

Worked solution

  1. Model the motion and choose the form of the acceleration

    a=vdvdxa=v\dfrac{dv}{dx}

    The force is a function of the displacement xx, so the useful form of the acceleration is v dv/dxv\,dv/dx.

  2. Write down Newton's second law for the particle PP

    mvdvdx=40−3x2mv\dfrac{dv}{dx}=40-3x^{2}

    The resultant force equals mass times acceleration.

  3. Separate the variables and integrate

    ∫3vmv dv=∫03(40−3x2)dx\int_{3}^{v} mv\,dv=\int_{0}^{3}\left(40-3x^{2}\right)dx

    Collecting vv on the left and xx on the right makes both sides integrable.

  4. Work out the work done by the force

    ∫03(40−3x2)dx=[−x3+40x]03=93 J\int_{0}^{3}\left(40-3x^{2}\right)dx=\left[- x^{3} + 40 x\right]_{0}^{3}=93\ \text{J}

    This definite integral is the work done by the variable force.

  5. Apply the work-energy principle

    12m(v2−32)=93\tfrac{1}{2}m\left(v^{2}-3^{2}\right)=93

    The work done equals the increase in kinetic energy.

  6. Recall the two forms of the acceleration

    a=dvdt=vdvdxa=\dfrac{dv}{dt}=v\dfrac{dv}{dx}

    The form to use depends on whether the force is given in terms of time or of displacement.

  7. Recall Newton's second law

    F=maF=ma

    The resultant force equals the mass times the acceleration at every instant.

  8. Recall the work done by a variable force

    W=∫F dxW=\int F\,dx

    The work of a force that varies with position is the integral of the force with respect to displacement.

  9. Recall the work-energy principle

    W=12mv2−12mu2W=\tfrac{1}{2}mv^{2}-\tfrac{1}{2}mu^{2}

    The work done by the resultant force equals the change in kinetic energy.

  10. Recall the impulse of a variable force

    J=∫F dtJ=\int F\,dt

    The impulse of a time-varying force is the integral of the force with respect to time.

  11. Recall the impulse-momentum principle

    J=mv−muJ=mv-mu

    The impulse of the resultant force equals the change in momentum.

  12. Recall that power is the rate of working

    P=FvP=Fv

    For a driving force FF at speed vv the power developed is FvFv.

  13. Recall the condition for the maximum speed

    a=0a=0

    At the maximum speed the acceleration is zero, so the driving force balances the resistance.

  14. Separate the variables before integrating

    ∫dvf(v)=∫dx\int \frac{dv}{f(v)}=\int dx

    A separable first-order equation is solved by collecting the speed on one side.

  15. State the value of gg used throughout

    g=9.8 m s−2g=9.8\ \text{m s}^{-2}

    All numerical answers use this value of the acceleration due to gravity.

  16. Solve for the required speed

    v=32+2×933=8.43 m s−1v=\sqrt{3^{2}+\dfrac{2\times 93}{3}}=8.43\ \text{m s}^{-1}

    Rearranging the work-energy equation gives the speed.

Answer
v=8.43 m s−1v=8.43\ \text{m s}^{-1}
Question 5
9 markschallenging
A car of mass 12001200 kg moves along a straight horizontal road. The engine of the car works at a constant rate of 1500015000 W. When the car is moving with speed vv m s−1^{-1} the total resistance to motion has magnitude 6v26v^{2} N. Find the acceleration of the car at the instant when its speed is 1212 m s−1^{-1}. Give your answer to 3 significant figures.
Show worked solution

Worked solution

  1. Relate the driving force to the power and the speed

    F=PvF=\dfrac{P}{v}

    The engine works at a constant rate, so the driving force is P/vP/v.

  2. Write down Newton's second law at this instant

    ma=Pv−6v2ma=\dfrac{P}{v}-6v^{2}

    The resultant force is the driving force minus the resistance.

  3. Evaluate the driving force and the resistance at this speed

    Pv=1500012=1250 N,6v2=864 N\dfrac{P}{v}=\dfrac{15000}{12}=1250\ \text{N},\quad 6v^{2}=864\ \text{N}

    Both the driving force and the resistance depend on the current speed.

  4. Recall the two forms of the acceleration

    a=dvdt=vdvdxa=\dfrac{dv}{dt}=v\dfrac{dv}{dx}

    The form to use depends on whether the force is given in terms of time or of displacement.

  5. Recall Newton's second law

    F=maF=ma

    The resultant force equals the mass times the acceleration at every instant.

  6. Recall the work done by a variable force

    W=∫F dxW=\int F\,dx

    The work of a force that varies with position is the integral of the force with respect to displacement.

  7. Recall the work-energy principle

    W=12mv2−12mu2W=\tfrac{1}{2}mv^{2}-\tfrac{1}{2}mu^{2}

    The work done by the resultant force equals the change in kinetic energy.

  8. Recall the impulse of a variable force

    J=∫F dtJ=\int F\,dt

    The impulse of a time-varying force is the integral of the force with respect to time.

  9. Recall the impulse-momentum principle

    J=mv−muJ=mv-mu

    The impulse of the resultant force equals the change in momentum.

  10. Recall that power is the rate of working

    P=FvP=Fv

    For a driving force FF at speed vv the power developed is FvFv.

  11. Recall the condition for the maximum speed

    a=0a=0

    At the maximum speed the acceleration is zero, so the driving force balances the resistance.

  12. Separate the variables before integrating

    ∫dvf(v)=∫dx\int \frac{dv}{f(v)}=\int dx

    A separable first-order equation is solved by collecting the speed on one side.

  13. State the value of gg used throughout

    g=9.8 m s−2g=9.8\ \text{m s}^{-2}

    All numerical answers use this value of the acceleration due to gravity.

  14. Check the units of each term

    [F]=N,[W]=J[F]=\text{N},\quad [W]=\text{J}

    Forces are in newtons and work and energy in joules.

  15. Solve for the acceleration

    a=1250−8641200=0.322 m s−2a=\dfrac{1250-864}{1200}=0.322\ \text{m s}^{-2}

    Dividing the resultant force by the mass gives the acceleration.

Answer
a=0.322 m s−2a=0.322\ \text{m s}^{-2}

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