Hard Further Maths Momentum and impulse Questions

Challenging, exam-style Further Maths Momentum and impulse questions with worked solutions. Stretch yourself on the hardest impulse-momentum-principle, impulse, sign-convention, conservation-of-momentum problems.

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Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
Two small spheres AA and BB, which are modelled as particles, are moving in the same straight line on a smooth horizontal plane. Take the direction of motion of AA immediately before the collision as the positive direction. The masses of AA and BB are mAm_{A} and mBm_{B}, their velocities immediately before the collision are uAu_{A} and uBu_{B}, and their velocities immediately afterwards are vAv_{A} and vBv_{B}. Which of the following expressions gives IBI_{B}, the impulse exerted on BB by AA during the collision?
Show worked solution

Worked solution

  1. State the impulse-momentum principle for BB

    IB=mBvB−mBuBI_{B}=m_{B}v_{B}-m_{B}u_{B}

    The impulse on BB equals the change in the momentum of BB.

  2. Factorise out the mass

    IB=mB(vB−uB)I_{B}=m_{B}\left(v_{B}-u_{B}\right)

    The mass is a common factor of the two momenta.

  3. State the positive direction being used

    positive→direction of motion of A before impact\text{positive}\rightarrow\text{direction of motion of }A\text{ before impact}

    Velocity and impulse are vectors, so a direction must be fixed before any numbers are written down.

  4. Note the equivalent expression from Newton's third law

    IB=−IA=−mA(vA−uA)I_{B}=-I_{A}=-m_{A}\left(v_{A}-u_{A}\right)

    Conservation of momentum makes the two expressions equal in value.

  5. Reject the option with the terms reversed

    mB(uB−vB)=−IBm_{B}\left(u_{B}-v_{B}\right)=-I_{B}

    Reversing the order of the two momenta reverses the direction of the impulse.

  6. Reject the option without a mass

    vB−uB≠IBv_{B}-u_{B}\neq I_{B}

    A change of velocity is not an impulse.

  7. State the impulse-momentum principle in vector form

    I=mv−mu\mathbf{I}=m\mathbf{v}-m\mathbf{u}

    The impulse acting on a body is equal to the change in its momentum.

  8. Recall that momentum is a vector

    p=mv\mathbf{p}=m\mathbf{v}

    Momentum points in the same direction as the velocity, so signs matter.

  9. Note the units of impulse and momentum

    1 N s=1 kg m s−11\ \text{N s}=1\ \text{kg m s}^{-1}

    Impulse and momentum have the same dimensions; either unit may be quoted.

  10. State Newton's third law for impulses

    IA=−IB\mathbf{I}_{A}=-\mathbf{I}_{B}

    The impulses the two bodies exert on each other are equal and opposite.

  11. Note why momentum is conserved here

    ∑Iexternal=0\sum\mathbf{I}_{\text{external}}=\mathbf{0}

    During the instantaneous interaction the only impulses are internal to the system.

  12. State the impulse of a constant force

    I=Ft\mathbf{I}=\mathbf{F}t

    A constant force acting for a time tt delivers an impulse Ft\mathbf{F}t.

  13. Record the modelling assumptions

    particles; smooth plane; instantaneous impact\text{particles; smooth plane; instantaneous impact}

    Modelling the bodies as particles removes rotation, and a smooth plane gives no frictional impulse.

  14. Check the sign convention

    positive=the direction stated in the question\text{positive}=\text{the direction stated in the question}

    Every velocity and every impulse must be measured with the same positive direction.

  15. Note that a negative answer is a physical result

    v<0  ⟺  motion in the negative directionv<0\iff\text{motion in the negative direction}

    A negative velocity simply means the body moves opposite to the chosen positive direction.

  16. Select the correct expression

    IB=mB(vB−uB)I_{B}=m_{B}\left(v_{B}-u_{B}\right)

    This is the only option that follows from the principle being used.

Answer
IB=mB(vB−uB)I_{B}=m_{B}\left(v_{B}-u_{B}\right)
Question 2
9 markschallenging
Two small spheres AA and BB, which are modelled as particles, are moving in the same straight line on a smooth horizontal plane. Take the direction of motion of AA immediately before the collision as the positive direction. The mass of AA is mA=3 kgm_{A}=3\ \text{kg} and the mass of BB is mB=2 kgm_{B}=2\ \text{kg}. Immediately before the collision the velocity of AA is uA=8 m s−1u_{A}=8\ \text{m s}^{-1} and the velocity of BB is uB=0 m s−1u_{B}=0\ \text{m s}^{-1}. The spheres collide. Immediately after the collision the velocity of AA is vA=2 m s−1v_{A}=2\ \text{m s}^{-1}. After the collision a constant force F=−3 NF=-3\ \text{N} acts on BB for a time t=4 st=4\ \text{s}. Which of the following is the value of wBw_{B}, the velocity of BB at the end of that time, in m s−1\text{m s}^{-1}?
Show worked solution

Worked solution

  1. State the principle of conservation of linear momentum

    mAuA+mBuB=mAvA+mBvBm_{A}u_{A}+m_{B}u_{B}=m_{A}v_{A}+m_{B}v_{B}

    The plane is smooth and the impact is instantaneous, so no external impulse acts on the system and the total momentum is unchanged.

  2. State the positive direction being used

    positive→direction of motion of A before impact\text{positive}\rightarrow\text{direction of motion of }A\text{ before impact}

    Velocity and impulse are vectors, so a direction must be fixed before any numbers are written down.

  3. Find the momentum of AA before the impact

    mAuA=3×8=24m_{A}u_{A}=3\times8=24

    Momentum is mass times velocity, keeping the sign of the velocity.

  4. Find the momentum of BB before the impact

    mBuB=2×0=0m_{B}u_{B}=2\times0=0

    The sign of uBu_{B} is carried through into the momentum.

  5. Find the total momentum before the impact

    pbefore=24+0=24p_{\text{before}}=24+0=24

    The two momenta are added, with their signs.

  6. Write the total momentum after the impact in terms of vBv_{B}

    pafter=3×2+2vBp_{\text{after}}=3\times2+2v_{B}

    Only vBv_{B} is unknown after the impact.

  7. Form the momentum equation

    24=6+2vB24=6+2v_{B}

    The momentum before equals the momentum after.

  8. Solve for vBv_{B}

    vB=24−62=9v_{B}=\frac{24-6}{2}=9

    Rearranging gives the velocity of BB immediately after the impact.

  9. Check the total momentum after the impact

    pafter=3×2+2×9=24p_{\text{after}}=3\times2+2\times9=24

    The two momenta after the impact are added with their signs.

  10. Confirm that momentum has been conserved

    pbefore=24=pafterp_{\text{before}}=24=p_{\text{after}}

    The total momentum before and after the impact agree, as they must.

  11. Find the impulse exerted on BB by AA

    IB=mB(vB−uB)=2(9−0)=18I_{B}=m_{B}\left(v_{B}-u_{B}\right)=2\left(9-0\right)=18

    The impulse on BB is its change in momentum.

  12. Find the impulse exerted on AA by BB

    IA=mA(vA−uA)=3(2−8)=−18I_{A}=m_{A}\left(v_{A}-u_{A}\right)=3\left(2-8\right)=-18

    The impulse on AA is its change in momentum.

  13. Check Newton's third law

    IA+IB=−18+18=0I_{A}+I_{B}=-18+18=0

    The two impulses are equal in magnitude and opposite in direction.

  14. Find the total kinetic energy before the impact

    Ekbefore=12×3×82+12×2×02=96E_{k}^{\text{before}}=\frac{1}{2}\times3\times8^{2}+\frac{1}{2}\times2\times0^{2}=96

    Kinetic energy is a scalar, so the signs of the velocities do not matter here.

  15. Select the option equal to this value

    wB=3 m s−1w_{B}=3\ \text{m s}^{-1}

    This is the required value, quoted with its units.

Answer
wB=3 m s−1w_{B}=3\ \text{m s}^{-1}
Question 3
9 markschallenging
Two small spheres AA and BB, which are modelled as particles, are moving in the same straight line on a smooth horizontal plane. Take the direction of motion of AA immediately before the collision as the positive direction. The mass of AA is mA=2 kgm_{A}=2\ \text{kg} and the mass of BB is mB=2 kgm_{B}=2\ \text{kg}. Immediately before the collision the velocity of AA is uA=6 m s−1u_{A}=6\ \text{m s}^{-1} and the velocity of BB is uB=−4 m s−1u_{B}=-4\ \text{m s}^{-1}. The spheres collide. Immediately after the collision the velocity of AA is vA=−2 m s−1v_{A}=-2\ \text{m s}^{-1}. Which of the following is the value of EE, the loss in kinetic energy of the system caused by the collision, in J\text{J}?
Show worked solution

Worked solution

  1. State the principle of conservation of linear momentum

    mAuA+mBuB=mAvA+mBvBm_{A}u_{A}+m_{B}u_{B}=m_{A}v_{A}+m_{B}v_{B}

    The plane is smooth and the impact is instantaneous, so no external impulse acts on the system and the total momentum is unchanged.

  2. State the positive direction being used

    positive→direction of motion of A before impact\text{positive}\rightarrow\text{direction of motion of }A\text{ before impact}

    Velocity and impulse are vectors, so a direction must be fixed before any numbers are written down.

  3. Find the momentum of AA before the impact

    mAuA=2×6=12m_{A}u_{A}=2\times6=12

    Momentum is mass times velocity, keeping the sign of the velocity.

  4. Find the momentum of BB before the impact

    mBuB=2×(−4)=−8m_{B}u_{B}=2\times\left(-4\right)=-8

    The sign of uBu_{B} is carried through into the momentum.

  5. Find the total momentum before the impact

    pbefore=12+(−8)=4p_{\text{before}}=12+\left(-8\right)=4

    The two momenta are added, with their signs.

  6. Write the total momentum after the impact in terms of vBv_{B}

    pafter=2×(−2)+2vBp_{\text{after}}=2\times\left(-2\right)+2v_{B}

    Only vBv_{B} is unknown after the impact.

  7. Form the momentum equation

    4=−4+2vB4=-4+2v_{B}

    The momentum before equals the momentum after.

  8. Solve for vBv_{B}

    vB=4−(−4)2=4v_{B}=\frac{4-\left(-4\right)}{2}=4

    Rearranging gives the velocity of BB immediately after the impact.

  9. Check the total momentum after the impact

    pafter=2×(−2)+2×4=4p_{\text{after}}=2\times\left(-2\right)+2\times4=4

    The two momenta after the impact are added with their signs.

  10. Confirm that momentum has been conserved

    pbefore=4=pafterp_{\text{before}}=4=p_{\text{after}}

    The total momentum before and after the impact agree, as they must.

  11. Find the impulse exerted on BB by AA

    IB=mB(vB−uB)=2(4−(−4))=16I_{B}=m_{B}\left(v_{B}-u_{B}\right)=2\left(4-\left(-4\right)\right)=16

    The impulse on BB is its change in momentum.

  12. Find the impulse exerted on AA by BB

    IA=mA(vA−uA)=2(−2−6)=−16I_{A}=m_{A}\left(v_{A}-u_{A}\right)=2\left(-2-6\right)=-16

    The impulse on AA is its change in momentum.

  13. Check Newton's third law

    IA+IB=−16+16=0I_{A}+I_{B}=-16+16=0

    The two impulses are equal in magnitude and opposite in direction.

  14. Find the total kinetic energy before the impact

    Ekbefore=12×2×62+12×2×(−4)2=52E_{k}^{\text{before}}=\frac{1}{2}\times2\times6^{2}+\frac{1}{2}\times2\times\left(-4\right)^{2}=52

    Kinetic energy is a scalar, so the signs of the velocities do not matter here.

  15. Find the total kinetic energy after the impact

    Ekafter=12×2×(−2)2+12×2×42=20E_{k}^{\text{after}}=\frac{1}{2}\times2\times\left(-2\right)^{2}+\frac{1}{2}\times2\times4^{2}=20

    The same formula is applied to the velocities after the impact.

  16. Confirm that no kinetic energy has been created

    20≤5220\leq52

    An impact can only destroy kinetic energy, so this is a valid check.

  17. Select the option equal to this value

    E=32 JE=32\ \text{J}

    This is the required value, quoted with its units.

Answer
E=32 JE=32\ \text{J}
Question 4
9 markschallenging
Two small spheres AA and BB, which are modelled as particles, are moving in the same straight line on a smooth horizontal plane. Take the direction of motion of AA immediately before the collision as the positive direction. The mass of AA is mA=2 kgm_{A}=2\ \text{kg} and the mass of BB is mB=5 kgm_{B}=5\ \text{kg}. Immediately before the collision the velocity of AA is uA=7 m s−1u_{A}=7\ \text{m s}^{-1} and the velocity of BB is uB=−1 m s−1u_{B}=-1\ \text{m s}^{-1}. The spheres collide. Immediately after the collision the velocity of AA is vA=−1 m s−1v_{A}=-1\ \text{m s}^{-1}. After the collision a constant force F=−2 NF=-2\ \text{N} acts on BB until its velocity is wB=0.2 m s−1w_{B}=0.2\ \text{m s}^{-1}. Find tt, the time for which the force acts, in s\text{s}.
Show worked solution

Worked solution

  1. State the principle of conservation of linear momentum

    mAuA+mBuB=mAvA+mBvBm_{A}u_{A}+m_{B}u_{B}=m_{A}v_{A}+m_{B}v_{B}

    The plane is smooth and the impact is instantaneous, so no external impulse acts on the system and the total momentum is unchanged.

  2. State the positive direction being used

    positive→direction of motion of A before impact\text{positive}\rightarrow\text{direction of motion of }A\text{ before impact}

    Velocity and impulse are vectors, so a direction must be fixed before any numbers are written down.

  3. Find the momentum of AA before the impact

    mAuA=2×7=14m_{A}u_{A}=2\times7=14

    Momentum is mass times velocity, keeping the sign of the velocity.

  4. Find the momentum of BB before the impact

    mBuB=5×(−1)=−5m_{B}u_{B}=5\times\left(-1\right)=-5

    The sign of uBu_{B} is carried through into the momentum.

  5. Find the total momentum before the impact

    pbefore=14+(−5)=9p_{\text{before}}=14+\left(-5\right)=9

    The two momenta are added, with their signs.

  6. Write the total momentum after the impact in terms of vBv_{B}

    pafter=2×(−1)+5vBp_{\text{after}}=2\times\left(-1\right)+5v_{B}

    Only vBv_{B} is unknown after the impact.

  7. Form the momentum equation

    9=−2+5vB9=-2+5v_{B}

    The momentum before equals the momentum after.

  8. Solve for vBv_{B}

    vB=9−(−2)5=2.2v_{B}=\frac{9-\left(-2\right)}{5}=2.2

    Rearranging gives the velocity of BB immediately after the impact.

  9. Check the total momentum after the impact

    pafter=2×(−1)+5×2.2=9p_{\text{after}}=2\times\left(-1\right)+5\times2.2=9

    The two momenta after the impact are added with their signs.

  10. Confirm that momentum has been conserved

    pbefore=9=pafterp_{\text{before}}=9=p_{\text{after}}

    The total momentum before and after the impact agree, as they must.

  11. Find the impulse exerted on BB by AA

    IB=mB(vB−uB)=5(2.2−(−1))=16I_{B}=m_{B}\left(v_{B}-u_{B}\right)=5\left(2.2-\left(-1\right)\right)=16

    The impulse on BB is its change in momentum.

  12. Find the impulse exerted on AA by BB

    IA=mA(vA−uA)=2(−1−7)=−16I_{A}=m_{A}\left(v_{A}-u_{A}\right)=2\left(-1-7\right)=-16

    The impulse on AA is its change in momentum.

  13. Check Newton's third law

    IA+IB=−16+16=0I_{A}+I_{B}=-16+16=0

    The two impulses are equal in magnitude and opposite in direction.

  14. Find the total kinetic energy before the impact

    Ekbefore=12×2×72+12×5×(−1)2=51.5E_{k}^{\text{before}}=\frac{1}{2}\times2\times7^{2}+\frac{1}{2}\times5\times\left(-1\right)^{2}=51.5

    Kinetic energy is a scalar, so the signs of the velocities do not matter here.

  15. Find the total kinetic energy after the impact

    Ekafter=12×2×(−1)2+12×5×2.22=13.1E_{k}^{\text{after}}=\frac{1}{2}\times2\times\left(-1\right)^{2}+\frac{1}{2}\times5\times2.2^{2}=13.1

    The same formula is applied to the velocities after the impact.

  16. State the final answer with its units

    t=5 st=5\ \text{s}

    This is the required value, quoted with its units.

Answer
t=5 st=5\ \text{s}
Question 5
9 markschallenging
A particle of mass m=3 kgm=3\ \text{kg} is moving on a smooth horizontal plane and is modelled as a particle. The vectors i\mathbf{i} and j\mathbf{j} are perpendicular unit vectors in a horizontal plane, and they fix the positive directions. Its velocity immediately before it receives an impulse is u=(2i−j) m s−1\mathbf{u}=\left(2\mathbf{i}-\mathbf{j}\right)\ \text{m s}^{-1} and its velocity immediately afterwards is v=(4i+3j) m s−1\mathbf{v}=\left(4\mathbf{i}+3\mathbf{j}\right)\ \text{m s}^{-1}. Find ∣I∣\left|\mathbf{I}\right|, the magnitude of the impulse received by the particle, in N s\text{N s}.
Show worked solution

Worked solution

  1. State the impulse-momentum principle in vector form

    I=mv−mu\mathbf{I}=m\mathbf{v}-m\mathbf{u}

    The impulse vector equals the change in the momentum vector.

  2. Note that the equation holds component by component

    i and j separately\mathbf{i}\ \text{and}\ \mathbf{j}\ \text{separately}

    The i\mathbf{i} and j\mathbf{j} components may be handled independently.

  3. Find the momentum before the impulse

    mu=3(2i−j)=6i−3jm\mathbf{u}=3\left(2\mathbf{i}-\mathbf{j}\right)=6\mathbf{i}-3\mathbf{j}

    Each component of the initial velocity is multiplied by the mass.

  4. Find the momentum after the impulse

    mv=3(4i+3j)=12i+9jm\mathbf{v}=3\left(4\mathbf{i}+3\mathbf{j}\right)=12\mathbf{i}+9\mathbf{j}

    Each component of the final velocity is multiplied by the mass.

  5. Subtract to find the impulse

    I=(12i+9j)−(6i−3j)=6i+12j\mathbf{I}=\left(12\mathbf{i}+9\mathbf{j}\right)-\left(6\mathbf{i}-3\mathbf{j}\right)=6\mathbf{i}+12\mathbf{j}

    The impulse is the change in momentum.

  6. Check the i\mathbf{i} component of the impulse

    Ix=3(4−2)=6I_{x}=3\left(4-2\right)=6

    The horizontal component of the change in momentum.

  7. Check the j\mathbf{j} component of the impulse

    Iy=3(3−(−1))=12I_{y}=3\left(3-\left(-1\right)\right)=12

    The vertical component of the change in momentum.

  8. Find the magnitude of the impulse

    ∣I∣=62+122=65\left|\mathbf{I}\right|=\sqrt{6^{2}+12^{2}}=6\sqrt{5}

    Pythagoras gives the magnitude of the impulse vector.

  9. Find the initial speed of the particle

    ∣u∣=22+(−1)2=5\left|\mathbf{u}\right|=\sqrt{2^{2}+\left(-1\right)^{2}}=\sqrt{5}

    The speed is the magnitude of the velocity vector.

  10. Find the final speed of the particle

    ∣v∣=42+32=5\left|\mathbf{v}\right|=\sqrt{4^{2}+3^{2}}=5

    Speed is a scalar and is always positive.

  11. State the impulse-momentum principle in vector form

    I=mv−mu\mathbf{I}=m\mathbf{v}-m\mathbf{u}

    The impulse acting on a body is equal to the change in its momentum.

  12. Recall that momentum is a vector

    p=mv\mathbf{p}=m\mathbf{v}

    Momentum points in the same direction as the velocity, so signs matter.

  13. Note the units of impulse and momentum

    1 N s=1 kg m s−11\ \text{N s}=1\ \text{kg m s}^{-1}

    Impulse and momentum have the same dimensions; either unit may be quoted.

  14. State Newton's third law for impulses

    IA=−IB\mathbf{I}_{A}=-\mathbf{I}_{B}

    The impulses the two bodies exert on each other are equal and opposite.

  15. State the final answer with its units

    ∣I∣=65 N s\left|\mathbf{I}\right|=6\sqrt{5}\ \text{N s}

    This is the required impulse, quoted with its units.

Answer
∣I∣=65 N s\left|\mathbf{I}\right|=6\sqrt{5}\ \text{N s}

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