Hard Further Maths Elastic collisions in two dimensions Questions

Challenging, exam-style Further Maths Elastic collisions in two dimensions questions with worked solutions. Stretch yourself on the hardest oblique-impact, line-of-centres, restitution, kinetic-energy-loss problems.

oblique-impactline-of-centresrestitutionkinetic-energy-lossimpulsedeflection-angle
Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
Two smooth spheres AA and BB, of masses mA=3m_{A}=3 kg and mB=4m_{B}=4 kg, are moving on a smooth horizontal surface and collide obliquely. At the instant of impact their velocities are resolved along the line of centres and perpendicular to it. The velocity of AA has components an=6a_{n}=6 m s−1^{-1} and at=5a_{t}=5 m s−1^{-1}, and the velocity of BB has components bn=−1b_{n}=-1 m s−1^{-1} and bt=2b_{t}=2 m s−1^{-1}. The coefficient of restitution between the spheres is e=12e=\frac{1}{2}. Take the positive direction along the line of centres to be the direction from AA towards BB, so that the approach speed along the line of centres is positive; the spheres are smooth, so the components perpendicular to the line of centres are unchanged. Which of the following correctly explains why the component of the velocity of AA perpendicular to the line of centres is unchanged, and gives its value?
Show worked solution

Worked solution

  1. State the reference directions and convention

    normal (n): along the line of centres, from A to B;tangential (t): perpendicular to it\text{normal }(n):\ \text{along the line of centres, from }A\text{ to }B;\quad\text{tangential }(t):\ \text{perpendicular to it}

    The positive direction along the line of centres is taken from AA towards BB, and every component below is signed accordingly.

  2. Write down the data for the impact

    mA=3, mB=4, an=6, at=5, bn=−1, bt=2, e=12m_{A}=3,\ m_{B}=4,\ a_{n}=6,\ a_{t}=5,\ b_{n}=-1,\ b_{t}=2,\ e=\frac{1}{2}

    Each velocity is resolved along the line of centres and perpendicular to it.

  3. Note why the impulse acts along the line of centres

    smooth spheres⇒no tangential impulse\text{smooth spheres}\Rightarrow\text{no tangential impulse}

    Because the spheres are smooth there is no friction at the contact, so the impulse is entirely along the line joining the centres.

  4. Carry the tangential component of AA through unchanged

    vAt=at=5v_{At}=a_{t}=5

    There is no impulse perpendicular to the line of centres, so this component of AA is unaffected by the impact.

  5. Carry the tangential component of BB through unchanged

    vBt=bt=2v_{Bt}=b_{t}=2

    For the same reason the tangential component of BB passes through the impact untouched.

  6. Write down conservation of momentum along the line of centres

    mAan+mBbn=mAvAn+mBvBnm_{A}a_{n}+m_{B}b_{n}=m_{A}v_{An}+m_{B}v_{Bn}

    Only the normal components change, so momentum is applied in that direction to relate them.

  7. Substitute the normal data into the momentum equation

    3(6)+4(−1)=3vAn+4vBn3\left(6\right)+4\left(-1\right)=3v_{An}+4v_{Bn}

    The signed normal components are inserted exactly as they stand.

  8. Simplify the momentum equation

    3vAn+4vBn=143v_{An}+4v_{Bn}=14

    This is the first of the two simultaneous equations for the normal components after impact.

  9. Write down Newton's law of restitution along the line of centres

    vBn−vAn=e(an−bn)v_{Bn}-v_{An}=e\left(a_{n}-b_{n}\right)

    Separation speed equals ee times approach speed, measured along the line of centres.

  10. Substitute the normal data into the restitution equation

    vBn−vAn=12(6−(−1))=72v_{Bn}-v_{An}=\frac{1}{2}\left(6-\left(-1\right)\right)=\frac{7}{2}

    The bracket is the approach speed along the line of centres.

  11. Solve the simultaneous equations for vBnv_{Bn}

    vBn=72v_{Bn}=\frac{7}{2}

    Adding mAm_{A} times the restitution equation to the momentum equation eliminates vAnv_{An}.

  12. Back-substitute to find vAnv_{An}

    vAn=vBn−72=0v_{An}=v_{Bn}-\frac{7}{2}=0

    The normal component of AA follows from the restitution equation.

  13. State the velocity of AA immediately after the impact

    vA=(0, 5)\mathbf{v}_{A}=\left(0,\ 5\right)

    The components are along the line of centres and perpendicular to it respectively.

  14. State the velocity of BB immediately after the impact

    vB=(72, 2)\mathbf{v}_{B}=\left(\frac{7}{2},\ 2\right)

    BB keeps its tangential component and takes the normal component found above.

  15. Find the speed of AA after the impact

    ∣vA∣=02+52=25≈5\left|\mathbf{v}_{A}\right|=\sqrt{0^{2}+5^{2}}=\sqrt{25}\approx5

    The resultant speed of AA combines its two components by Pythagoras' theorem.

  16. Find the speed of BB after the impact

    ∣vB∣=722+22=654≈4.03\left|\mathbf{v}_{B}\right|=\sqrt{\frac{7}{2}^{2}+2^{2}}=\sqrt{\frac{65}{4}}\approx4.03

    The resultant speed of BB is found in the same way.

Answer
vAt=5v_{At}=5, because the spheres are smooth, so there is no impulse perpendicular to the line of centres and the tangential component of AA is unchanged.
Question 2
9 markschallenging
Two smooth spheres AA and BB, of masses mA=3m_{A}=3 kg and mB=3m_{B}=3 kg, are moving on a smooth ice rink and collide obliquely. At the instant of impact their velocities are resolved along the line of centres and perpendicular to it. The velocity of AA has components an=7a_{n}=7 m s−1^{-1} and at=4a_{t}=4 m s−1^{-1}, and the velocity of BB has components bn=0b_{n}=0 m s−1^{-1} and bt=0b_{t}=0 m s−1^{-1}. The coefficient of restitution between the spheres is e=1e=1. Take the positive direction along the line of centres to be the direction from AA towards BB, so that the approach speed along the line of centres is positive; the spheres are smooth, so the components perpendicular to the line of centres are unchanged. Which of the following correctly describes the motion of AA immediately after this perfectly elastic impact between equal masses?
Show worked solution

Worked solution

  1. State the reference directions and convention

    normal (n): along the line of centres, from A to B;tangential (t): perpendicular to it\text{normal }(n):\ \text{along the line of centres, from }A\text{ to }B;\quad\text{tangential }(t):\ \text{perpendicular to it}

    The positive direction along the line of centres is taken from AA towards BB, and every component below is signed accordingly.

  2. Write down the data for the impact

    mA=3, mB=3, an=7, at=4, bn=0, bt=0, e=1m_{A}=3,\ m_{B}=3,\ a_{n}=7,\ a_{t}=4,\ b_{n}=0,\ b_{t}=0,\ e=1

    Each velocity is resolved along the line of centres and perpendicular to it.

  3. Note why the impulse acts along the line of centres

    smooth spheres⇒no tangential impulse\text{smooth spheres}\Rightarrow\text{no tangential impulse}

    Because the spheres are smooth there is no friction at the contact, so the impulse is entirely along the line joining the centres.

  4. Carry the tangential component of AA through unchanged

    vAt=at=4v_{At}=a_{t}=4

    There is no impulse perpendicular to the line of centres, so this component of AA is unaffected by the impact.

  5. Carry the tangential component of BB through unchanged

    vBt=bt=0v_{Bt}=b_{t}=0

    For the same reason the tangential component of BB passes through the impact untouched.

  6. Write down conservation of momentum along the line of centres

    mAan+mBbn=mAvAn+mBvBnm_{A}a_{n}+m_{B}b_{n}=m_{A}v_{An}+m_{B}v_{Bn}

    Only the normal components change, so momentum is applied in that direction to relate them.

  7. Substitute the normal data into the momentum equation

    3(7)+3(0)=3vAn+3vBn3\left(7\right)+3\left(0\right)=3v_{An}+3v_{Bn}

    The signed normal components are inserted exactly as they stand.

  8. Simplify the momentum equation

    3vAn+3vBn=213v_{An}+3v_{Bn}=21

    This is the first of the two simultaneous equations for the normal components after impact.

  9. Write down Newton's law of restitution along the line of centres

    vBn−vAn=e(an−bn)v_{Bn}-v_{An}=e\left(a_{n}-b_{n}\right)

    Separation speed equals ee times approach speed, measured along the line of centres.

  10. Substitute the normal data into the restitution equation

    vBn−vAn=1(7−0)=7v_{Bn}-v_{An}=1\left(7-0\right)=7

    The bracket is the approach speed along the line of centres.

  11. Solve the simultaneous equations for vBnv_{Bn}

    vBn=7v_{Bn}=7

    Adding mAm_{A} times the restitution equation to the momentum equation eliminates vAnv_{An}.

  12. Back-substitute to find vAnv_{An}

    vAn=vBn−7=0v_{An}=v_{Bn}-7=0

    The normal component of AA follows from the restitution equation.

  13. State the velocity of AA immediately after the impact

    vA=(0, 4)\mathbf{v}_{A}=\left(0,\ 4\right)

    The components are along the line of centres and perpendicular to it respectively.

  14. State the velocity of BB immediately after the impact

    vB=(7, 0)\mathbf{v}_{B}=\left(7,\ 0\right)

    BB keeps its tangential component and takes the normal component found above.

  15. Check momentum along the line of centres

    3(0)+3(7)=213\left(0\right)+3\left(7\right)=21

    The total normal momentum after the impact equals its value before, as it must.

Answer
vAn=0v_{An}=0, so AA moves off along the perpendicular to the line of centres only: equal masses with e=1e=1 exchange their line-of-centres components, and BB was at rest along that line.
Question 3
9 markschallenging
Two smooth spheres AA and BB, of masses mA=7m_{A}=7 kg and mB=2m_{B}=2 kg, are moving on a smooth horizontal floor and collide obliquely. At the instant of impact their velocities are resolved along the line of centres and perpendicular to it. The velocity of AA has components an=6a_{n}=6 m s−1^{-1} and at=3a_{t}=3 m s−1^{-1}, and the velocity of BB has components bn=−1b_{n}=-1 m s−1^{-1} and bt=1b_{t}=1 m s−1^{-1}. The coefficient of restitution between the spheres is e=12e=\frac{1}{2}. Take the positive direction along the line of centres to be the direction from AA towards BB, so that the approach speed along the line of centres is positive; the spheres are smooth, so the components perpendicular to the line of centres are unchanged. Find the magnitude of the impulse exerted on BB by AA.
Show worked solution

Worked solution

  1. State the reference directions and convention

    normal (n): along the line of centres, from A to B;tangential (t): perpendicular to it\text{normal }(n):\ \text{along the line of centres, from }A\text{ to }B;\quad\text{tangential }(t):\ \text{perpendicular to it}

    The positive direction along the line of centres is taken from AA towards BB, and every component below is signed accordingly.

  2. Write down the data for the impact

    mA=7, mB=2, an=6, at=3, bn=−1, bt=1, e=12m_{A}=7,\ m_{B}=2,\ a_{n}=6,\ a_{t}=3,\ b_{n}=-1,\ b_{t}=1,\ e=\frac{1}{2}

    Each velocity is resolved along the line of centres and perpendicular to it.

  3. Note why the impulse acts along the line of centres

    smooth spheres⇒no tangential impulse\text{smooth spheres}\Rightarrow\text{no tangential impulse}

    Because the spheres are smooth there is no friction at the contact, so the impulse is entirely along the line joining the centres.

  4. Carry the tangential component of AA through unchanged

    vAt=at=3v_{At}=a_{t}=3

    There is no impulse perpendicular to the line of centres, so this component of AA is unaffected by the impact.

  5. Carry the tangential component of BB through unchanged

    vBt=bt=1v_{Bt}=b_{t}=1

    For the same reason the tangential component of BB passes through the impact untouched.

  6. Write down conservation of momentum along the line of centres

    mAan+mBbn=mAvAn+mBvBnm_{A}a_{n}+m_{B}b_{n}=m_{A}v_{An}+m_{B}v_{Bn}

    Only the normal components change, so momentum is applied in that direction to relate them.

  7. Substitute the normal data into the momentum equation

    7(6)+2(−1)=7vAn+2vBn7\left(6\right)+2\left(-1\right)=7v_{An}+2v_{Bn}

    The signed normal components are inserted exactly as they stand.

  8. Simplify the momentum equation

    7vAn+2vBn=407v_{An}+2v_{Bn}=40

    This is the first of the two simultaneous equations for the normal components after impact.

  9. Write down Newton's law of restitution along the line of centres

    vBn−vAn=e(an−bn)v_{Bn}-v_{An}=e\left(a_{n}-b_{n}\right)

    Separation speed equals ee times approach speed, measured along the line of centres.

  10. Substitute the normal data into the restitution equation

    vBn−vAn=12(6−(−1))=72v_{Bn}-v_{An}=\frac{1}{2}\left(6-\left(-1\right)\right)=\frac{7}{2}

    The bracket is the approach speed along the line of centres.

  11. Solve the simultaneous equations for vBnv_{Bn}

    vBn=436v_{Bn}=\frac{43}{6}

    Adding mAm_{A} times the restitution equation to the momentum equation eliminates vAnv_{An}.

  12. Back-substitute to find vAnv_{An}

    vAn=vBn−72=113v_{An}=v_{Bn}-\frac{7}{2}=\frac{11}{3}

    The normal component of AA follows from the restitution equation.

  13. State the velocity of AA immediately after the impact

    vA=(113, 3)\mathbf{v}_{A}=\left(\frac{11}{3},\ 3\right)

    The components are along the line of centres and perpendicular to it respectively.

  14. State the velocity of BB immediately after the impact

    vB=(436, 1)\mathbf{v}_{B}=\left(\frac{43}{6},\ 1\right)

    BB keeps its tangential component and takes the normal component found above.

  15. Find the speed of AA after the impact

    ∣vA∣=1132+32=2029≈4.74\left|\mathbf{v}_{A}\right|=\sqrt{\frac{11}{3}^{2}+3^{2}}=\sqrt{\frac{202}{9}}\approx4.74

    The resultant speed of AA combines its two components by Pythagoras' theorem.

  16. Find the speed of BB after the impact

    ∣vB∣=4362+12=188536≈7.24\left|\mathbf{v}_{B}\right|=\sqrt{\frac{43}{6}^{2}+1^{2}}=\sqrt{\frac{1885}{36}}\approx7.24

    The resultant speed of BB is found in the same way.

  17. Write down the impulse on BB along the line of centres

    I=mB(vBn−bn)I=m_{B}\left(v_{Bn}-b_{n}\right)

    The impulse on BB equals its change of normal momentum; there is no tangential impulse.

  18. Evaluate the magnitude of the impulse between the spheres

    ∣I∣=2(436−(−1))=493\left|I\right|=2\left(\frac{43}{6}-\left(-1\right)\right)=\frac{49}{3}

    The impulse on AA is equal in magnitude and opposite in direction.

Answer
∣I∣=493 N s\left|I\right|=\frac{49}{3}\text{ N s}
Question 4
9 markschallenging
Two smooth spheres AA and BB, of masses mA=1m_{A}=1 kg and mB=1m_{B}=1 kg, are moving on a smooth horizontal plane and collide obliquely. At the instant of impact their velocities are resolved along the line of centres and perpendicular to it. The velocity of AA has components an=8a_{n}=8 m s−1^{-1} and at=4a_{t}=4 m s−1^{-1}, and the velocity of BB has components bn=0b_{n}=0 m s−1^{-1} and bt=0b_{t}=0 m s−1^{-1}. The coefficient of restitution between the spheres is e=1e=1. Take the positive direction along the line of centres to be the direction from AA towards BB, so that the approach speed along the line of centres is positive; the spheres are smooth, so the components perpendicular to the line of centres are unchanged. Find the angle through which the direction of motion of AA is deflected by the impact.
Show worked solution

Worked solution

  1. State the reference directions and convention

    normal (n): along the line of centres, from A to B;tangential (t): perpendicular to it\text{normal }(n):\ \text{along the line of centres, from }A\text{ to }B;\quad\text{tangential }(t):\ \text{perpendicular to it}

    The positive direction along the line of centres is taken from AA towards BB, and every component below is signed accordingly.

  2. Write down the data for the impact

    mA=1, mB=1, an=8, at=4, bn=0, bt=0, e=1m_{A}=1,\ m_{B}=1,\ a_{n}=8,\ a_{t}=4,\ b_{n}=0,\ b_{t}=0,\ e=1

    Each velocity is resolved along the line of centres and perpendicular to it.

  3. Note why the impulse acts along the line of centres

    smooth spheres⇒no tangential impulse\text{smooth spheres}\Rightarrow\text{no tangential impulse}

    Because the spheres are smooth there is no friction at the contact, so the impulse is entirely along the line joining the centres.

  4. Carry the tangential component of AA through unchanged

    vAt=at=4v_{At}=a_{t}=4

    There is no impulse perpendicular to the line of centres, so this component of AA is unaffected by the impact.

  5. Carry the tangential component of BB through unchanged

    vBt=bt=0v_{Bt}=b_{t}=0

    For the same reason the tangential component of BB passes through the impact untouched.

  6. Write down conservation of momentum along the line of centres

    mAan+mBbn=mAvAn+mBvBnm_{A}a_{n}+m_{B}b_{n}=m_{A}v_{An}+m_{B}v_{Bn}

    Only the normal components change, so momentum is applied in that direction to relate them.

  7. Substitute the normal data into the momentum equation

    1(8)+1(0)=vAn+vBn1\left(8\right)+1\left(0\right)=v_{An}+v_{Bn}

    The signed normal components are inserted exactly as they stand.

  8. Simplify the momentum equation

    vAn+vBn=8v_{An}+v_{Bn}=8

    This is the first of the two simultaneous equations for the normal components after impact.

  9. Write down Newton's law of restitution along the line of centres

    vBn−vAn=e(an−bn)v_{Bn}-v_{An}=e\left(a_{n}-b_{n}\right)

    Separation speed equals ee times approach speed, measured along the line of centres.

  10. Substitute the normal data into the restitution equation

    vBn−vAn=1(8−0)=8v_{Bn}-v_{An}=1\left(8-0\right)=8

    The bracket is the approach speed along the line of centres.

  11. Solve the simultaneous equations for vBnv_{Bn}

    vBn=8v_{Bn}=8

    Adding mAm_{A} times the restitution equation to the momentum equation eliminates vAnv_{An}.

  12. Back-substitute to find vAnv_{An}

    vAn=vBn−8=0v_{An}=v_{Bn}-8=0

    The normal component of AA follows from the restitution equation.

  13. State the velocity of AA immediately after the impact

    vA=(0, 4)\mathbf{v}_{A}=\left(0,\ 4\right)

    The components are along the line of centres and perpendicular to it respectively.

  14. State the velocity of BB immediately after the impact

    vB=(8, 0)\mathbf{v}_{B}=\left(8,\ 0\right)

    BB keeps its tangential component and takes the normal component found above.

  15. Find the speed of AA after the impact

    ∣vA∣=02+42=16≈4\left|\mathbf{v}_{A}\right|=\sqrt{0^{2}+4^{2}}=\sqrt{16}\approx4

    The resultant speed of AA combines its two components by Pythagoras' theorem.

  16. Find the angle through which AA is deflected

    cos⁡α=uA⋅vA∣uA∣∣vA∣ ⇒ α≈63.4∘\cos\alpha=\frac{\mathbf{u}_{A}\cdot\mathbf{v}_{A}}{\left|\mathbf{u}_{A}\right|\left|\mathbf{v}_{A}\right|}\ \Rightarrow\ \alpha\approx63.4^{\circ}

    The deflection is the angle between the velocity of AA before and after the impact, found from the scalar product.

Answer
α≈63.4∘\alpha\approx63.4^{\circ}
Question 5
9 markschallenging
A smooth sphere of mass m=2m=2 kg moves on a smooth horizontal table between two parallel fixed smooth vertical walls. Resolved perpendicular to the walls and parallel to them, its velocity immediately before its first impact has components un=9u_{n}=9 m s−1^{-1} (towards the first wall) and ut=6u_{t}=6 m s−1^{-1} (parallel to the walls). Take the positive perpendicular direction to be the direction of the sphere's approach to the first wall. The coefficient of restitution between the sphere and each wall is e=13e=\frac{1}{3}. The sphere strikes the first wall, rebounds, crosses to the second wall and strikes it. Which of the following gives the components of its velocity immediately after the second impact?
Show worked solution

Worked solution

  1. State the reference directions and convention

    normal (n): perpendicular to the walls;tangential (t): parallel to the walls\text{normal }(n):\ \text{perpendicular to the walls};\quad\text{tangential }(t):\ \text{parallel to the walls}

    The positive normal direction is the direction of the sphere's approach to the first wall; the parallel component is common to both walls.

  2. Write down the data for the motion

    m=2,un=9,ut=6,e=13m=2,\quad u_{n}=9,\quad u_{t}=6,\quad e=\frac{1}{3}

    The parallel component is unchanged at every impact; only the perpendicular component is affected.

  3. Carry the parallel component through both impacts

    vt=ut=6v_{t}=u_{t}=6

    Each wall is smooth, so neither impact exerts an impulse along the walls and the parallel component is constant throughout.

  4. Reverse and scale the normal component at the first wall

    n1=−e un=−3n_{1}=-e\,u_{n}=-3

    The first impact multiplies the perpendicular component by ee and reverses it, sending the sphere towards the second wall.

  5. State the velocity after the first impact

    v1=(−3, 6)\mathbf{v}_{1}=\left(-3,\ 6\right)

    The perpendicular component is now directed towards the second wall while the parallel component is unchanged.

  6. Reverse and scale the normal component at the second wall

    n2=−e n1=e2un=1n_{2}=-e\,n_{1}=e^{2}u_{n}=1

    The second impact multiplies the perpendicular component by ee again, so after two impacts it is e2e^{2} times the original.

  7. State the velocity immediately after the second impact

    v=(1, 6)\mathbf{v}=\left(1,\ 6\right)

    After two impacts the perpendicular component has magnitude e2une^{2}u_{n} and the parallel component is still utu_{t}.

  8. Find the speed after the second impact

    ∣v∣=12+62=37≈6.08\left|\mathbf{v}\right|=\sqrt{1^{2}+6^{2}}=\sqrt{37}\approx6.08

    The resultant speed combines the reduced perpendicular component with the unchanged parallel component.

  9. Find the angle to the walls after the second impact

    tan⁡θ=16 ⇒ θ≈9.46∘\tan\theta=\frac{1}{6}\ \Rightarrow\ \theta\approx9.46^{\circ}

    The path becomes more nearly parallel to the walls as the perpendicular component is repeatedly reduced.

  10. Find the kinetic energy before the impacts

    KEbefore=12(2)(92+62)=117\text{KE}_{\text{before}}=\frac{1}{2}\left(2\right)\left(9^{2}+6^{2}\right)=117

    This is the kinetic energy of the sphere as it approaches the first wall.

  11. Find the kinetic energy after the second impact

    KEafter=12(2)(12+62)=37\text{KE}_{\text{after}}=\frac{1}{2}\left(2\right)\left(1^{2}+6^{2}\right)=37

    Only the perpendicular part of the kinetic energy has been reduced, by a factor e4e^{4}.

  12. Find the total kinetic energy lost in the two impacts

    KE lost=117−37=12mun2(1−e4)=80\text{KE lost}=117-37=\frac{1}{2}mu_{n}^{2}\left(1-e^{4}\right)=80

    The total loss is the sum of the losses at the two walls, and depends only on the perpendicular motion.

  13. Find the impulse at the first wall

    ∣I1∣=m un(1+e)=24\left|I_{1}\right|=m\,u_{n}\left(1+e\right)=24

    The first wall reverses a perpendicular component of magnitude unu_{n}.

  14. Find the impulse at the second wall

    ∣I2∣=m e un(1+e)=8\left|I_{2}\right|=m\,e\,u_{n}\left(1+e\right)=8

    The second wall reverses the smaller perpendicular component e une\,u_{n}.

  15. Note why the spheres are modelled as smooth

    no tangential impulse⇒impulse acts along the line of centres\text{no tangential impulse}\Rightarrow\text{impulse acts along the line of centres}

    Smooth spheres exert no friction on each other at contact, so the whole impulse is directed along the line joining their centres.

  16. Check the pattern of the perpendicular component

    ∣nk∣=ekun\left|n_{k}\right|=e^{k}u_{n}

    After kk impacts the perpendicular component has magnitude ekune^{k}u_{n}, which shrinks geometrically.

Answer
v=(1, 6) m s−1\mathbf{v}=\left(1,\ 6\right)\text{ m s}^{-1}

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