Hard Further Maths Elastic collisions in one dimension Questions

Challenging, exam-style Further Maths Elastic collisions in one dimension questions with worked solutions. Stretch yourself on the hardest restitution, kinetic-energy-loss, direct-impact, impulse problems.

restitutionkinetic-energy-lossdirect-impactimpulsecoefficient-of-restitutionsuccessive-collisions
Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
Sphere BB, of mass mB=4m_{B}=4 kg, is at rest on a smooth horizontal plane between a fixed smooth vertical wall and a sphere AA of mass mA=5m_{A}=5 kg. The centres of the two smooth spheres lie on a straight line perpendicular to the wall. Take the positive direction to be the direction in which AA is moving immediately before the impact. Immediately before the first impact the velocities are uA=7u_{A}=7 m s−1^{-1} and uB=0u_{B}=0 m s−1^{-1}, so AA is moving towards BB and the wall. The coefficient of restitution between the two spheres is e=25e=\frac{2}{5}, and the coefficient of restitution between BB and the wall is ew=12e_{w}=\frac{1}{2}. After the first impact BB travels to the wall, rebounds from it, and then collides with AA for a second time. The wall is fixed, so it is effectively of infinite mass and conservation of momentum must not be applied to the sphere on its own. Which of the following correctly gives the separation speed of AA and BB immediately after their second collision, and explains what it means?
Show worked solution

Worked solution

  1. Write down the data for the impact (first collision of AA with BB)

    mA=5,mB=4,uA=7,uB=0,e=25m_{A}=5,\quad m_{B}=4,\quad u_{A}=7,\quad u_{B}=0,\quad e=\frac{2}{5}

    Masses are in kilograms and velocities in metres per second, each signed with respect to the positive direction.

  2. Substitute the data into the momentum equation (first collision of AA with BB)

    5(7)+4(0)=5vA+4vB5\left(7\right)+4\left(0\right)=5v_{A}+4v_{B}

    The signed velocities are put in exactly as they stand, negatives included.

  3. Simplify the momentum equation (first collision of AA with BB)

    5vA+4vB=355v_{A}+4v_{B}=35

    This is the first of the two simultaneous equations for the unknown velocities.

  4. Substitute the data into the restitution equation (first collision of AA with BB)

    vB−vA=−25(0−7)v_{B}-v_{A}=-\frac{2}{5}\left(0-7\right)

    The bracket is the approach velocity uB−uAu_{B}-u_{A}, and the minus sign in front of ee is what reverses it into a separation.

  5. Simplify the restitution equation (first collision of AA with BB)

    vB−vA=145v_{B}-v_{A}=\frac{14}{5}

    This is the second simultaneous equation; note that the separation speed is never negative.

  6. Solve for vBv_{B} (first collision of AA with BB)

    vB=499v_{B}=\frac{49}{9}

    Dividing by the total mass gives the velocity of B immediately after the impact.

  7. Back-substitute to find vAv_{A} (first collision of AA with BB)

    vA=499−145=11945v_{A}=\frac{49}{9}-\frac{14}{5}=\frac{119}{45}

    Putting the value of vBv_{B} back into the restitution equation gives the velocity of A.

  8. State the velocities immediately after the impact (first collision of AA with BB)

    vA=11945,vB=499v_{A}=\frac{119}{45},\quad v_{B}=\frac{49}{9}

    Both velocities are positive, so both spheres continue in the positive direction.

  9. Apply the law of restitution at the wall

    wB=−ewvB=−12×499w_{B}=-e_{w}v_{B}=-\frac{1}{2}\times\frac{49}{9}

    The sphere leaves the wall with ewe_{w} times the speed at which it arrived, but in the opposite direction.

  10. State the velocity of BB after the wall

    wB=−4918w_{B}=-\frac{49}{18}

    The velocity is negative, so BB is now travelling back towards AA.

  11. Check that BB does catch AA

    vA−wB=11945−(−4918)=16130>0v_{A}-w_{B}=\frac{119}{45}-\left(-\frac{49}{18}\right)=\frac{161}{30}>0

    BB lies between AA and the wall and is now moving in the negative direction, while AA is still moving towards it; the approach speed is positive, so a second collision does happen.

  12. Write down the data for the impact (second collision of AA with BB)

    mA=5,mB=4,vA=11945,wB=−4918,e=25m_{A}=5,\quad m_{B}=4,\quad v_{A}=\frac{119}{45},\quad w_{B}=-\frac{49}{18},\quad e=\frac{2}{5}

    Masses are in kilograms and velocities in metres per second, each signed with respect to the positive direction.

  13. Substitute the data into the momentum equation (second collision of AA with BB)

    5(11945)+4(−4918)=5xA+4xB5\left(\frac{119}{45}\right)+4\left(-\frac{49}{18}\right)=5x_{A}+4x_{B}

    The signed velocities are put in exactly as they stand, negatives included.

  14. Simplify the momentum equation (second collision of AA with BB)

    5xA+4xB=735x_{A}+4x_{B}=\frac{7}{3}

    This is the first of the two simultaneous equations for the unknown velocities.

  15. Substitute the data into the restitution equation (second collision of AA with BB)

    xB−xA=−25(−4918−11945)x_{B}-x_{A}=-\frac{2}{5}\left(-\frac{49}{18}-\frac{119}{45}\right)

    The bracket is the approach velocity wB−vAw_{B}-v_{A}, and the minus sign in front of ee is what reverses it into a separation.

  16. Simplify the restitution equation (second collision of AA with BB)

    xB−xA=16175x_{B}-x_{A}=\frac{161}{75}

    This is the second simultaneous equation; note that the separation speed is never negative.

  17. State the velocities immediately after the impact (second collision of AA with BB)

    xA=−469675,xB=196135x_{A}=-\frac{469}{675},\quad x_{B}=\frac{196}{135}

    A negative value means that sphere moves in the negative direction, which is a genuine physical result.

Answer
xB−xA=16175x_{B}-x_{A}=\frac{161}{75} m s−1^{-1}, the separation speed, which is e=25e=\frac{2}{5} times the approach speed 16130\frac{161}{30} m s−1^{-1} of the second collision; because it is not negative, AA can never catch BB again.
Question 2
9 markschallenging
Sphere BB, of mass mB=2m_{B}=2 kg, is at rest on a smooth horizontal table between a fixed smooth vertical wall and a sphere AA of mass mA=3m_{A}=3 kg. The centres of the two smooth spheres lie on a straight line perpendicular to the wall. Take the positive direction to be the direction in which AA is moving immediately before the impact. Immediately before the first impact the velocities are uA=8u_{A}=8 m s−1^{-1} and uB=0u_{B}=0 m s−1^{-1}, so AA is moving towards BB and the wall. The coefficient of restitution between the two spheres is e=12e=\frac{1}{2}, and the coefficient of restitution between BB and the wall is ew=23e_{w}=\frac{2}{3}. After the first impact BB travels to the wall, rebounds from it, and then collides with AA for a second time. The wall is fixed, so it is effectively of infinite mass and conservation of momentum must not be applied to the sphere on its own. Which of the following correctly gives the separation speed of AA and BB immediately after their second collision, and explains what it means?
Show worked solution

Worked solution

  1. Write down the data for the impact (first collision of AA with BB)

    mA=3,mB=2,uA=8,uB=0,e=12m_{A}=3,\quad m_{B}=2,\quad u_{A}=8,\quad u_{B}=0,\quad e=\frac{1}{2}

    Masses are in kilograms and velocities in metres per second, each signed with respect to the positive direction.

  2. Substitute the data into the momentum equation (first collision of AA with BB)

    3(8)+2(0)=3vA+2vB3\left(8\right)+2\left(0\right)=3v_{A}+2v_{B}

    The signed velocities are put in exactly as they stand, negatives included.

  3. Simplify the momentum equation (first collision of AA with BB)

    3vA+2vB=243v_{A}+2v_{B}=24

    This is the first of the two simultaneous equations for the unknown velocities.

  4. Substitute the data into the restitution equation (first collision of AA with BB)

    vB−vA=−12(0−8)v_{B}-v_{A}=-\frac{1}{2}\left(0-8\right)

    The bracket is the approach velocity uB−uAu_{B}-u_{A}, and the minus sign in front of ee is what reverses it into a separation.

  5. Simplify the restitution equation (first collision of AA with BB)

    vB−vA=4v_{B}-v_{A}=4

    This is the second simultaneous equation; note that the separation speed is never negative.

  6. Solve for vBv_{B} (first collision of AA with BB)

    vB=365v_{B}=\frac{36}{5}

    Dividing by the total mass gives the velocity of B immediately after the impact.

  7. Back-substitute to find vAv_{A} (first collision of AA with BB)

    vA=365−4=165v_{A}=\frac{36}{5}-4=\frac{16}{5}

    Putting the value of vBv_{B} back into the restitution equation gives the velocity of A.

  8. State the velocities immediately after the impact (first collision of AA with BB)

    vA=165,vB=365v_{A}=\frac{16}{5},\quad v_{B}=\frac{36}{5}

    Both velocities are positive, so both spheres continue in the positive direction.

  9. Apply the law of restitution at the wall

    wB=−ewvB=−23×365w_{B}=-e_{w}v_{B}=-\frac{2}{3}\times\frac{36}{5}

    The sphere leaves the wall with ewe_{w} times the speed at which it arrived, but in the opposite direction.

  10. State the velocity of BB after the wall

    wB=−245w_{B}=-\frac{24}{5}

    The velocity is negative, so BB is now travelling back towards AA.

  11. Check that BB does catch AA

    vA−wB=165−(−245)=8>0v_{A}-w_{B}=\frac{16}{5}-\left(-\frac{24}{5}\right)=8>0

    BB lies between AA and the wall and is now moving in the negative direction, while AA is still moving towards it; the approach speed is positive, so a second collision does happen.

  12. Write down the data for the impact (second collision of AA with BB)

    mA=3,mB=2,vA=165,wB=−245,e=12m_{A}=3,\quad m_{B}=2,\quad v_{A}=\frac{16}{5},\quad w_{B}=-\frac{24}{5},\quad e=\frac{1}{2}

    Masses are in kilograms and velocities in metres per second, each signed with respect to the positive direction.

  13. Substitute the data into the momentum equation (second collision of AA with BB)

    3(165)+2(−245)=3xA+2xB3\left(\frac{16}{5}\right)+2\left(-\frac{24}{5}\right)=3x_{A}+2x_{B}

    The signed velocities are put in exactly as they stand, negatives included.

  14. Simplify the momentum equation (second collision of AA with BB)

    3xA+2xB=03x_{A}+2x_{B}=0

    This is the first of the two simultaneous equations for the unknown velocities.

  15. Simplify the restitution equation (second collision of AA with BB)

    xB−xA=4x_{B}-x_{A}=4

    This is the second simultaneous equation; note that the separation speed is never negative.

  16. State the velocities immediately after the impact (second collision of AA with BB)

    xA=−85,xB=125x_{A}=-\frac{8}{5},\quad x_{B}=\frac{12}{5}

    A negative value means that sphere moves in the negative direction, which is a genuine physical result.

Answer
xB−xA=4x_{B}-x_{A}=4 m s−1^{-1}, the separation speed, which is e=12e=\frac{1}{2} times the approach speed 88 m s−1^{-1} of the second collision; because it is not negative, AA can never catch BB again.
Question 3
9 markschallenging
Sphere BB, of mass mB=2m_{B}=2 kg, is at rest on a smooth horizontal surface between a fixed smooth vertical wall and a sphere AA of mass mA=6m_{A}=6 kg. The centres of the two smooth spheres lie on a straight line perpendicular to the wall. Take the positive direction to be the direction in which AA is moving immediately before the impact. Immediately before the first impact the velocities are uA=5u_{A}=5 m s−1^{-1} and uB=0u_{B}=0 m s−1^{-1}, so AA is moving towards BB and the wall. The coefficient of restitution between the two spheres is e=12e=\frac{1}{2}, and the coefficient of restitution between BB and the wall is ew=12e_{w}=\frac{1}{2}. After the first impact BB travels to the wall, rebounds from it, and then collides with AA for a second time. The wall is fixed, so it is effectively of infinite mass and conservation of momentum must not be applied to the sphere on its own. Which of the following is the total kinetic energy lost in the three impacts?
Show worked solution

Worked solution

  1. Write down the data for the impact (first collision of AA with BB)

    mA=6,mB=2,uA=5,uB=0,e=12m_{A}=6,\quad m_{B}=2,\quad u_{A}=5,\quad u_{B}=0,\quad e=\frac{1}{2}

    Masses are in kilograms and velocities in metres per second, each signed with respect to the positive direction.

  2. Substitute the data into the momentum equation (first collision of AA with BB)

    6(5)+2(0)=6vA+2vB6\left(5\right)+2\left(0\right)=6v_{A}+2v_{B}

    The signed velocities are put in exactly as they stand, negatives included.

  3. Simplify the momentum equation (first collision of AA with BB)

    6vA+2vB=306v_{A}+2v_{B}=30

    This is the first of the two simultaneous equations for the unknown velocities.

  4. Substitute the data into the restitution equation (first collision of AA with BB)

    vB−vA=−12(0−5)v_{B}-v_{A}=-\frac{1}{2}\left(0-5\right)

    The bracket is the approach velocity uB−uAu_{B}-u_{A}, and the minus sign in front of ee is what reverses it into a separation.

  5. Simplify the restitution equation (first collision of AA with BB)

    vB−vA=52v_{B}-v_{A}=\frac{5}{2}

    This is the second simultaneous equation; note that the separation speed is never negative.

  6. Solve for vBv_{B} (first collision of AA with BB)

    vB=458v_{B}=\frac{45}{8}

    Dividing by the total mass gives the velocity of B immediately after the impact.

  7. Back-substitute to find vAv_{A} (first collision of AA with BB)

    vA=458−52=258v_{A}=\frac{45}{8}-\frac{5}{2}=\frac{25}{8}

    Putting the value of vBv_{B} back into the restitution equation gives the velocity of A.

  8. State the velocities immediately after the impact (first collision of AA with BB)

    vA=258,vB=458v_{A}=\frac{25}{8},\quad v_{B}=\frac{45}{8}

    Both velocities are positive, so both spheres continue in the positive direction.

  9. Find the total kinetic energy before the impact (first collision of AA with BB)

    KEbefore=12(6)(5)2+12(2)(0)2=75\text{KE}_{\text{before}}=\frac{1}{2}\left(6\right)\left(5\right)^{2}+\frac{1}{2}\left(2\right)\left(0\right)^{2}=75

    Kinetic energy depends on the square of the speed, so the signs of the velocities do not matter here.

  10. Apply the law of restitution at the wall

    wB=−ewvB=−12×458w_{B}=-e_{w}v_{B}=-\frac{1}{2}\times\frac{45}{8}

    The sphere leaves the wall with ewe_{w} times the speed at which it arrived, but in the opposite direction.

  11. State the velocity of BB after the wall

    wB=−4516w_{B}=-\frac{45}{16}

    The velocity is negative, so BB is now travelling back towards AA.

  12. Check that BB does catch AA

    vA−wB=258−(−4516)=9516>0v_{A}-w_{B}=\frac{25}{8}-\left(-\frac{45}{16}\right)=\frac{95}{16}>0

    BB lies between AA and the wall and is now moving in the negative direction, while AA is still moving towards it; the approach speed is positive, so a second collision does happen.

  13. Write down the data for the impact (second collision of AA with BB)

    mA=6,mB=2,vA=258,wB=−4516,e=12m_{A}=6,\quad m_{B}=2,\quad v_{A}=\frac{25}{8},\quad w_{B}=-\frac{45}{16},\quad e=\frac{1}{2}

    Masses are in kilograms and velocities in metres per second, each signed with respect to the positive direction.

  14. Simplify the momentum equation (second collision of AA with BB)

    6xA+2xB=10586x_{A}+2x_{B}=\frac{105}{8}

    This is the first of the two simultaneous equations for the unknown velocities.

  15. Simplify the restitution equation (second collision of AA with BB)

    xB−xA=9532x_{B}-x_{A}=\frac{95}{32}

    This is the second simultaneous equation; note that the separation speed is never negative.

  16. State the velocities immediately after the impact (second collision of AA with BB)

    xA=115128,xB=495128x_{A}=\frac{115}{128},\quad x_{B}=\frac{495}{128}

    Both velocities are positive, so both spheres continue in the positive direction.

  17. Find the total kinetic energy at the very start

    KEstart=12(6)(5)2+12(2)(0)2=75\text{KE}_{\text{start}}=\frac{1}{2}\left(6\right)\left(5\right)^{2}+\frac{1}{2}\left(2\right)\left(0\right)^{2}=75

    The surface is smooth, so kinetic energy only changes at the impacts.

  18. Find the total kinetic energy at the very end

    KEend=12(6)(115128)2+12(2)(495128)2=711754096\text{KE}_{\text{end}}=\frac{1}{2}\left(6\right)\left(\frac{115}{128}\right)^{2}+\frac{1}{2}\left(2\right)\left(\frac{495}{128}\right)^{2}=\frac{71175}{4096}

    This uses the velocities immediately after the second collision between the spheres.

  19. Find the total kinetic energy lost in the three impacts

    KE lost=75−711754096=2360254096\text{KE lost}=75-\frac{71175}{4096}=\frac{236025}{4096}

    This equals the sum of the losses in the three separate impacts, 22516+6075256+812254096\frac{225}{16}+\frac{6075}{256}+\frac{81225}{4096}.

Answer
2360254096\frac{236025}{4096}
Question 4
9 markschallenging
Sphere BB, of mass mB=3m_{B}=3 kg, is at rest on a smooth horizontal floor between a fixed smooth vertical wall and a sphere AA of mass mA=5m_{A}=5 kg. The centres of the two smooth spheres lie on a straight line perpendicular to the wall. Take the positive direction to be the direction in which AA is moving immediately before the impact. Immediately before the first impact the velocities are uA=6u_{A}=6 m s−1^{-1} and uB=0u_{B}=0 m s−1^{-1}, so AA is moving towards BB and the wall. The coefficient of restitution between the two spheres is e=25e=\frac{2}{5}, and the coefficient of restitution between BB and the wall is ew=35e_{w}=\frac{3}{5}. After the first impact BB travels to the wall, rebounds from it, and then collides with AA for a second time. The wall is fixed, so it is effectively of infinite mass and conservation of momentum must not be applied to the sphere on its own. Which of the following gives the velocities of AA and BB immediately after their second collision with each other?
Show worked solution

Worked solution

  1. Write down the data for the impact (first collision of AA with BB)

    mA=5,mB=3,uA=6,uB=0,e=25m_{A}=5,\quad m_{B}=3,\quad u_{A}=6,\quad u_{B}=0,\quad e=\frac{2}{5}

    Masses are in kilograms and velocities in metres per second, each signed with respect to the positive direction.

  2. Substitute the data into the momentum equation (first collision of AA with BB)

    5(6)+3(0)=5vA+3vB5\left(6\right)+3\left(0\right)=5v_{A}+3v_{B}

    The signed velocities are put in exactly as they stand, negatives included.

  3. Simplify the momentum equation (first collision of AA with BB)

    5vA+3vB=305v_{A}+3v_{B}=30

    This is the first of the two simultaneous equations for the unknown velocities.

  4. Substitute the data into the restitution equation (first collision of AA with BB)

    vB−vA=−25(0−6)v_{B}-v_{A}=-\frac{2}{5}\left(0-6\right)

    The bracket is the approach velocity uB−uAu_{B}-u_{A}, and the minus sign in front of ee is what reverses it into a separation.

  5. Simplify the restitution equation (first collision of AA with BB)

    vB−vA=125v_{B}-v_{A}=\frac{12}{5}

    This is the second simultaneous equation; note that the separation speed is never negative.

  6. Solve for vBv_{B} (first collision of AA with BB)

    vB=428=214v_{B}=\frac{42}{8}=\frac{21}{4}

    Dividing by the total mass gives the velocity of B immediately after the impact.

  7. Back-substitute to find vAv_{A} (first collision of AA with BB)

    vA=214−125=5720v_{A}=\frac{21}{4}-\frac{12}{5}=\frac{57}{20}

    Putting the value of vBv_{B} back into the restitution equation gives the velocity of A.

  8. State the velocities immediately after the impact (first collision of AA with BB)

    vA=5720,vB=214v_{A}=\frac{57}{20},\quad v_{B}=\frac{21}{4}

    Both velocities are positive, so both spheres continue in the positive direction.

  9. Apply the law of restitution at the wall

    wB=−ewvB=−35×214w_{B}=-e_{w}v_{B}=-\frac{3}{5}\times\frac{21}{4}

    The sphere leaves the wall with ewe_{w} times the speed at which it arrived, but in the opposite direction.

  10. State the velocity of BB after the wall

    wB=−6320w_{B}=-\frac{63}{20}

    The velocity is negative, so BB is now travelling back towards AA.

  11. Check that BB does catch AA

    vA−wB=5720−(−6320)=6>0v_{A}-w_{B}=\frac{57}{20}-\left(-\frac{63}{20}\right)=6>0

    BB lies between AA and the wall and is now moving in the negative direction, while AA is still moving towards it; the approach speed is positive, so a second collision does happen.

  12. Write down the data for the impact (second collision of AA with BB)

    mA=5,mB=3,vA=5720,wB=−6320,e=25m_{A}=5,\quad m_{B}=3,\quad v_{A}=\frac{57}{20},\quad w_{B}=-\frac{63}{20},\quad e=\frac{2}{5}

    Masses are in kilograms and velocities in metres per second, each signed with respect to the positive direction.

  13. Substitute the data into the momentum equation (second collision of AA with BB)

    5(5720)+3(−6320)=5xA+3xB5\left(\frac{57}{20}\right)+3\left(-\frac{63}{20}\right)=5x_{A}+3x_{B}

    The signed velocities are put in exactly as they stand, negatives included.

  14. Simplify the momentum equation (second collision of AA with BB)

    5xA+3xB=2455x_{A}+3x_{B}=\frac{24}{5}

    This is the first of the two simultaneous equations for the unknown velocities.

  15. Substitute the data into the restitution equation (second collision of AA with BB)

    xB−xA=−25(−6320−5720)x_{B}-x_{A}=-\frac{2}{5}\left(-\frac{63}{20}-\frac{57}{20}\right)

    The bracket is the approach velocity wB−vAw_{B}-v_{A}, and the minus sign in front of ee is what reverses it into a separation.

  16. Simplify the restitution equation (second collision of AA with BB)

    xB−xA=125x_{B}-x_{A}=\frac{12}{5}

    This is the second simultaneous equation; note that the separation speed is never negative.

  17. State the velocities immediately after the impact (second collision of AA with BB)

    xA=−310,xB=2110x_{A}=-\frac{3}{10},\quad x_{B}=\frac{21}{10}

    A negative value means that sphere moves in the negative direction, which is a genuine physical result.

Answer
xA=−310,xB=2110x_{A}=-\frac{3}{10},\quad x_{B}=\frac{21}{10}
Question 5
9 markschallenging
Sphere BB, of mass mB=3m_{B}=3 kg, is at rest on a smooth horizontal plane between a fixed smooth vertical wall and a sphere AA of mass mA=4m_{A}=4 kg. The centres of the two smooth spheres lie on a straight line perpendicular to the wall. Take the positive direction to be the direction in which AA is moving immediately before the impact. Immediately before the first impact the velocities are uA=9u_{A}=9 m s−1^{-1} and uB=0u_{B}=0 m s−1^{-1}, so AA is moving towards BB and the wall. The coefficient of restitution between the two spheres is e=13e=\frac{1}{3}, and the coefficient of restitution between BB and the wall is ew=12e_{w}=\frac{1}{2}. After the first impact BB travels to the wall, rebounds from it, and then collides with AA for a second time. The wall is fixed, so it is effectively of infinite mass and conservation of momentum must not be applied to the sphere on its own. Which of the following gives the velocities of AA and BB immediately after their second collision with each other?
Show worked solution

Worked solution

  1. Write down the data for the impact (first collision of AA with BB)

    mA=4,mB=3,uA=9,uB=0,e=13m_{A}=4,\quad m_{B}=3,\quad u_{A}=9,\quad u_{B}=0,\quad e=\frac{1}{3}

    Masses are in kilograms and velocities in metres per second, each signed with respect to the positive direction.

  2. Substitute the data into the momentum equation (first collision of AA with BB)

    4(9)+3(0)=4vA+3vB4\left(9\right)+3\left(0\right)=4v_{A}+3v_{B}

    The signed velocities are put in exactly as they stand, negatives included.

  3. Simplify the momentum equation (first collision of AA with BB)

    4vA+3vB=364v_{A}+3v_{B}=36

    This is the first of the two simultaneous equations for the unknown velocities.

  4. Substitute the data into the restitution equation (first collision of AA with BB)

    vB−vA=−13(0−9)v_{B}-v_{A}=-\frac{1}{3}\left(0-9\right)

    The bracket is the approach velocity uB−uAu_{B}-u_{A}, and the minus sign in front of ee is what reverses it into a separation.

  5. Simplify the restitution equation (first collision of AA with BB)

    vB−vA=3v_{B}-v_{A}=3

    This is the second simultaneous equation; note that the separation speed is never negative.

  6. Solve for vBv_{B} (first collision of AA with BB)

    vB=487v_{B}=\frac{48}{7}

    Dividing by the total mass gives the velocity of B immediately after the impact.

  7. Back-substitute to find vAv_{A} (first collision of AA with BB)

    vA=487−3=277v_{A}=\frac{48}{7}-3=\frac{27}{7}

    Putting the value of vBv_{B} back into the restitution equation gives the velocity of A.

  8. State the velocities immediately after the impact (first collision of AA with BB)

    vA=277,vB=487v_{A}=\frac{27}{7},\quad v_{B}=\frac{48}{7}

    Both velocities are positive, so both spheres continue in the positive direction.

  9. Apply the law of restitution at the wall

    wB=−ewvB=−12×487w_{B}=-e_{w}v_{B}=-\frac{1}{2}\times\frac{48}{7}

    The sphere leaves the wall with ewe_{w} times the speed at which it arrived, but in the opposite direction.

  10. State the velocity of BB after the wall

    wB=−247w_{B}=-\frac{24}{7}

    The velocity is negative, so BB is now travelling back towards AA.

  11. Check that BB does catch AA

    vA−wB=277−(−247)=517>0v_{A}-w_{B}=\frac{27}{7}-\left(-\frac{24}{7}\right)=\frac{51}{7}>0

    BB lies between AA and the wall and is now moving in the negative direction, while AA is still moving towards it; the approach speed is positive, so a second collision does happen.

  12. Write down the data for the impact (second collision of AA with BB)

    mA=4,mB=3,vA=277,wB=−247,e=13m_{A}=4,\quad m_{B}=3,\quad v_{A}=\frac{27}{7},\quad w_{B}=-\frac{24}{7},\quad e=\frac{1}{3}

    Masses are in kilograms and velocities in metres per second, each signed with respect to the positive direction.

  13. Substitute the data into the momentum equation (second collision of AA with BB)

    4(277)+3(−247)=4xA+3xB4\left(\frac{27}{7}\right)+3\left(-\frac{24}{7}\right)=4x_{A}+3x_{B}

    The signed velocities are put in exactly as they stand, negatives included.

  14. Simplify the momentum equation (second collision of AA with BB)

    4xA+3xB=3674x_{A}+3x_{B}=\frac{36}{7}

    This is the first of the two simultaneous equations for the unknown velocities.

  15. Simplify the restitution equation (second collision of AA with BB)

    xB−xA=177x_{B}-x_{A}=\frac{17}{7}

    This is the second simultaneous equation; note that the separation speed is never negative.

  16. State the velocities immediately after the impact (second collision of AA with BB)

    xA=−1549,xB=10449x_{A}=-\frac{15}{49},\quad x_{B}=\frac{104}{49}

    A negative value means that sphere moves in the negative direction, which is a genuine physical result.

Answer
xA=−1549,xB=10449x_{A}=-\frac{15}{49},\quad x_{B}=\frac{104}{49}

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