Further Maths Elastic collisions in two dimensions Practice Questions

Free Further Maths Elastic collisions in two dimensions practice questions with full step-by-step worked solutions. Covers oblique-impact, impact-with-a-fixed-plane, restitution, kinetic-energy-loss. Practise exam-style problems and check your method.

oblique-impactimpact-with-a-fixed-planerestitutionkinetic-energy-lossimpulsecoefficient-of-restitution
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
A smooth sphere of mass m=12m=\frac{1}{2} kg is sliding on a smooth horizontal table and strikes a fixed smooth vertical wall obliquely. Resolved parallel to the wall and perpendicular to it, its velocity immediately before the impact has components ut=4u_{t}=4 m s−1^{-1} (parallel to the wall) and un=3u_{n}=3 m s−1^{-1} (towards the wall). Take the positive perpendicular direction to be the direction of the sphere's approach, so the perpendicular component of velocity is reversed by the impact while the component parallel to the wall is unchanged. The coefficient of restitution between the sphere and the wall is e=12e=\frac{1}{2}. Find the components of its velocity immediately after the impact with the wall.
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Worked solution

  1. Write down the data for the impact

    m=12,un=3,ut=4,e=12m=\frac{1}{2},\quad u_{n}=3,\quad u_{t}=4,\quad e=\frac{1}{2}

    The velocity is resolved into a component towards the wall and a component along it, both in metres per second.

  2. Carry the tangential component through unchanged

    vt=ut=4v_{t}=u_{t}=4

    The wall is smooth, so it exerts no impulse along its own surface and the parallel component of velocity is unaltered.

  3. State the velocity immediately after the impact

    v=(−32, 4)\mathbf{v}=\left(-\frac{3}{2},\ 4\right)

    The first entry is the component perpendicular to the wall (now directed away from it) and the second is the unchanged component along the wall.

Answer
v=(−32, 4) m s−1\mathbf{v}=\left(-\frac{3}{2},\ 4\right)\text{ m s}^{-1}
Question 2
2 markseasy
A smooth sphere of mass m=1m=1 kg is sliding on a smooth horizontal floor and strikes a fixed smooth vertical wall obliquely. Resolved parallel to the wall and perpendicular to it, its velocity immediately before the impact has components ut=5u_{t}=5 m s−1^{-1} (parallel to the wall) and un=10u_{n}=10 m s−1^{-1} (towards the wall). Take the positive perpendicular direction to be the direction of the sphere's approach, so the perpendicular component of velocity is reversed by the impact while the component parallel to the wall is unchanged. Immediately after the impact the component of its velocity perpendicular to the wall has magnitude vn=4v_{n}=4 m s−1^{-1}. Find the coefficient of restitution between the sphere and the wall.
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Worked solution

  1. Write down the data

    m=1,un=10,ut=5,vn=4m=1,\quad u_{n}=10,\quad u_{t}=5,\quad v_{n}=4

    The magnitude of the perpendicular component after the impact is given, and ee is the unknown.

  2. State the velocity after the impact with its correct sign

    v=(−4, 5)\mathbf{v}=\left(-4,\ 5\right)

    The perpendicular component is directed away from the wall, so it is negative with respect to the direction of approach.

  3. Substitute the two perpendicular components

    e=410=25e=\frac{4}{10}=\frac{2}{5}

    The parallel component plays no part in the coefficient of restitution.

Answer
e=25e=\frac{2}{5}
Question 3
4 marksintermediate
A smooth sphere of mass m=2m=2 kg is sliding on a smooth horizontal table and strikes a fixed smooth vertical wall obliquely. Resolved parallel to the wall and perpendicular to it, its velocity immediately before the impact has components ut=8u_{t}=8 m s−1^{-1} (parallel to the wall) and un=6u_{n}=6 m s−1^{-1} (towards the wall). Take the positive perpendicular direction to be the direction of the sphere's approach, so the perpendicular component of velocity is reversed by the impact while the component parallel to the wall is unchanged. The coefficient of restitution between the sphere and the wall is e=12e=\frac{1}{2}. Which of the following correctly explains why the component of the sphere's velocity parallel to the wall is unchanged, and gives its value?
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Worked solution

  1. State the reference directions and convention

    normal (n): towards the wall;tangential (t): along the wall\text{normal }(n):\ \text{towards the wall};\quad\text{tangential }(t):\ \text{along the wall}

    The positive normal direction is the direction of approach, so the impact reverses the normal component and leaves the tangential component alone.

  2. Write down the data for the impact

    m=2,un=6,ut=8,e=12m=2,\quad u_{n}=6,\quad u_{t}=8,\quad e=\frac{1}{2}

    The velocity is resolved into a component towards the wall and a component along it, both in metres per second.

  3. Carry the tangential component through unchanged

    vt=ut=8v_{t}=u_{t}=8

    The wall is smooth, so it exerts no impulse along its own surface and the parallel component of velocity is unaltered.

  4. Reverse and scale the normal component

    vn=−e un=−12×6=−3v_{n}=-e\,u_{n}=-\frac{1}{2}\times6=-3

    Newton's law of restitution applied perpendicular to the wall multiplies the approach component by ee and reverses its direction.

  5. State the velocity immediately after the impact

    v=(−3, 8)\mathbf{v}=\left(-3,\ 8\right)

    The first entry is the component perpendicular to the wall (now directed away from it) and the second is the unchanged component along the wall.

  6. Find the speed immediately after the impact

    ∣v∣=−32+82=73≈8.54\left|\mathbf{v}\right|=\sqrt{-3^{2}+8^{2}}=\sqrt{73}\approx8.54

    The resultant speed combines the two perpendicular components by Pythagoras' theorem.

  7. Find the angle the rebound makes with the wall

    tan⁡θ=∣vn∣vt=38 ⇒ θ≈20.6∘\tan\theta=\frac{\left|v_{n}\right|}{v_{t}}=\frac{3}{8}\ \Rightarrow\ \theta\approx20.6^{\circ}

    The angle to the wall has tangent equal to the perpendicular component over the parallel component.

Answer
vt=8v_{t}=8, because the wall is smooth, so it exerts no impulse along its own surface and the parallel component of velocity is unchanged.
Question 4
6 markshard
Two smooth spheres AA and BB, of masses mA=2m_{A}=2 kg and mB=3m_{B}=3 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=2a_{t}=2 m s−1^{-1}, and the velocity of BB has components bn=−2b_{n}=-2 m s−1^{-1} and bt=1b_{t}=1 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 gives the kinetic energy lost in this perfectly elastic impact, with the reason?
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Worked solution

  1. Write down the data for the impact

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

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

  2. Carry the tangential component of AA through unchanged

    vAt=at=2v_{At}=a_{t}=2

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

  3. 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.

  4. Substitute the normal data into the momentum equation

    2(6)+3(−2)=2vAn+3vBn2\left(6\right)+3\left(-2\right)=2v_{An}+3v_{Bn}

    The signed normal components are inserted exactly as they stand.

  5. Simplify the momentum equation

    2vAn+3vBn=62v_{An}+3v_{Bn}=6

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

  6. Substitute the normal data into the restitution equation

    vBn−vAn=1(6−(−2))=8v_{Bn}-v_{An}=1\left(6-\left(-2\right)\right)=8

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

  7. Solve the simultaneous equations for vBnv_{Bn}

    vBn=225v_{Bn}=\frac{22}{5}

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

  8. State the velocity of AA immediately after the impact

    vA=(−185, 2)\mathbf{v}_{A}=\left(-\frac{18}{5},\ 2\right)

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

  9. State the velocity of BB immediately after the impact

    vB=(225, 1)\mathbf{v}_{B}=\left(\frac{22}{5},\ 1\right)

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

  10. Find the total kinetic energy before the impact

    KEbefore=12(2)(62+22)+12(3)(−22+12)=952\text{KE}_{\text{before}}=\frac{1}{2}\left(2\right)\left(6^{2}+2^{2}\right)+\frac{1}{2}\left(3\right)\left(-2^{2}+1^{2}\right)=\frac{95}{2}

    Each sphere contributes the kinetic energy of the resultant of its two components.

  11. Find the total kinetic energy after the impact

    KEafter=12(2)(−1852+22)+12(3)(2252+12)=952\text{KE}_{\text{after}}=\frac{1}{2}\left(2\right)\left(-\frac{18}{5}^{2}+2^{2}\right)+\frac{1}{2}\left(3\right)\left(\frac{22}{5}^{2}+1^{2}\right)=\frac{95}{2}

    The same calculation is repeated with the velocities found after the impact.

  12. Find the kinetic energy lost in the impact

    KE lost=952−952=0\text{KE lost}=\frac{95}{2}-\frac{95}{2}=0

    The loss comes entirely from the change in the line-of-centres motion.

Answer
KE lost=0\text{KE lost}=0, because e=1e=1 makes the impact perfectly elastic, so the total kinetic energy after the impact equals the total kinetic energy before it.
Question 5
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.

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