Further Maths Elastic collisions in one dimension Practice Questions

Free Further Maths Elastic collisions in one dimension practice questions with full step-by-step worked solutions. Covers restitution, impact-with-a-fixed-plane, conservation-of-momentum, direct-impact. Practise exam-style problems and check your method.

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Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
A smooth sphere of mass m=12m=\frac{1}{2} kg slides on a smooth horizontal table and strikes a fixed smooth vertical wall at right angles. Take the positive direction to be the direction in which the sphere is moving immediately before it strikes the wall. 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. Immediately before the impact its velocity is u=6u=6 m s−1^{-1} and the coefficient of restitution between the sphere and the wall is e=23e=\frac{2}{3}. Find the velocity of the sphere immediately after the impact with the wall.
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Worked solution

  1. Write down the data for the impact with the wall

    m=12,u=6,e=23m=\frac{1}{2},\quad u=6,\quad e=\frac{2}{3}

    The wall is fixed, so its mass is effectively infinite and only the sphere has a velocity to track.

  2. Apply Newton's law of restitution at the wall

    rebound speed=e×approach speed\text{rebound speed}=e\times\text{approach speed}

    For a fixed plane the law reduces to this single statement; no momentum equation is needed.

  3. State the velocity of the sphere after the impact

    v=−4v=-4

    The direction of motion is reversed by the wall, so the velocity is negative with respect to the chosen positive direction.

Answer
v=−4 m s−1v=-4\text{ m s}^{-1}
Question 2
2 markseasy
Two smooth spheres AA and BB, of masses mA=2m_{A}=2 kg and mB=5m_{B}=5 kg, move on a smooth horizontal plane. Both spheres are moving in the same direction along the line joining their centres, with AA behind BB and catching it up. Take the positive direction to be the direction in which AA is moving immediately before the impact. Immediately before the impact their velocities are uA=6u_{A}=6 m s−1^{-1} and uB=1u_{B}=1 m s−1^{-1}, and the coefficient of restitution between the spheres is e=34e=\frac{3}{4}. Which of the following gives the velocities of AA and BB immediately after the impact?
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Worked solution

  1. Write down the data for the impact

    mA=2,mB=5,uA=6,uB=1,e=34m_{A}=2,\quad m_{B}=5,\quad u_{A}=6,\quad u_{B}=1,\quad e=\frac{3}{4}

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

  2. Simplify the momentum equation

    2vA+5vB=172v_{A}+5v_{B}=17

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

  3. Simplify the restitution equation

    vB−vA=154v_{B}-v_{A}=\frac{15}{4}

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

  4. State the velocities immediately after the impact

    vA=−14,vB=72v_{A}=-\frac{1}{4},\quad v_{B}=\frac{7}{2}

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

Answer
vA=−14,vB=72v_{A}=-\frac{1}{4},\quad v_{B}=\frac{7}{2}
Question 3
4 marksintermediate
A smooth sphere of mass m=2m=2 kg slides on a smooth horizontal surface and strikes a fixed smooth vertical wall at right angles. Take the positive direction to be the direction in which the sphere is moving immediately before it strikes the wall. 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. Immediately before the impact its velocity is u=6u=6 m s−1^{-1} and the coefficient of restitution between the sphere and the wall is e=13e=\frac{1}{3}. Which of the following correctly gives the magnitude of the impulse exerted on the sphere by the wall, and explains why the momentum of the sphere alone is not conserved?
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Worked solution

  1. Write down the data for the impact with the wall

    m=2,u=6,e=13m=2,\quad u=6,\quad e=\frac{1}{3}

    The wall is fixed, so its mass is effectively infinite and only the sphere has a velocity to track.

  2. Apply Newton's law of restitution at the wall

    rebound speed=e×approach speed\text{rebound speed}=e\times\text{approach speed}

    For a fixed plane the law reduces to this single statement; no momentum equation is needed.

  3. Substitute the data

    ∣v∣=13×6=2\left|v\right|=\frac{1}{3}\times6=2

    This is the speed with which the sphere leaves the wall.

  4. State the velocity of the sphere after the impact

    v=−2v=-2

    The direction of motion is reversed by the wall, so the velocity is negative with respect to the chosen positive direction.

  5. Write down the impulse-momentum principle for the sphere

    I=mv−muI=mv-mu

    The wall exerts an external impulse on the sphere, which is why its momentum changes.

  6. Substitute the values

    I=2(−2)−2(6)=−16I=2\left(-2\right)-2\left(6\right)=-16

    The impulse is negative because it acts away from the wall, in the negative direction.

  7. State the magnitude of the impulse

    ∣I∣=2(6)(1+13)=16\left|I\right|=2\left(6\right)\left(1+\frac{1}{3}\right)=16

    The magnitude of the impulse from a fixed plane is mu(1+e)mu\left(1+e\right) newton seconds.

Answer
Impulse=16\text{Impulse}=16 N s, because the wall is fixed to the Earth and so exerts an external impulse mu(1+e)m u\left(1+e\right) on the sphere; the momentum of the sphere alone is therefore not conserved.
Question 4
6 markshard
Two smooth spheres AA and BB, of masses mA=4m_{A}=4 kg and mB=2m_{B}=2 kg, move on a smooth horizontal table. Sphere BB is at rest and sphere AA is moving directly towards it along the line joining their centres. Take the positive direction to be the direction in which AA is moving immediately before the impact. Immediately before the impact their velocities are uA=9u_{A}=9 m s−1^{-1} and uB=0u_{B}=0 m s−1^{-1}, and the coefficient of restitution between the spheres is e=0e=0. Which of the following correctly describes the motion of the spheres immediately after the impact?
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Worked solution

  1. Write down the data for the impact

    mA=4,mB=2,uA=9,uB=0,e=0m_{A}=4,\quad m_{B}=2,\quad u_{A}=9,\quad u_{B}=0,\quad e=0

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

  2. Write down conservation of linear momentum

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

    There is no external horizontal impulse on the pair of spheres, so their total momentum is unchanged by the impact.

  3. Substitute the data into the momentum equation

    4(9)+2(0)=4vA+2vB4\left(9\right)+2\left(0\right)=4v_{A}+2v_{B}

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

  4. Simplify the momentum equation

    4vA+2vB=364v_{A}+2v_{B}=36

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

  5. Write down Newton's law of restitution

    vB−vA=−e(uB−uA)v_{B}-v_{A}=-e\left(u_{B}-u_{A}\right)

    Separation speed equals ee times approach speed; with one fixed positive direction this is the correct signed form.

  6. Substitute the data into the restitution equation

    vB−vA=−0(0−9)v_{B}-v_{A}=-0\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.

  7. Simplify the restitution equation

    vB−vA=0v_{B}-v_{A}=0

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

  8. Solve for vBv_{B}

    vB=366=6v_{B}=\frac{36}{6}=6

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

  9. Deduce vAv_{A}

    vA=vB=6v_{A}=v_{B}=6

    The spheres coalesce, so A has the same velocity as B.

  10. State the velocities immediately after the impact

    vA=6,vB=6v_{A}=6,\quad v_{B}=6

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

Answer
vA=6v_{A}=6 m s−1^{-1}, and vBv_{B} takes the same value: with e=0e=0 the separation speed is zero, so the spheres coalesce and move on together.
Question 5
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?
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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.

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