Elastic collisions in one dimension Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Elastic collisions in one dimension questions. See exactly how to solve problems on restitution, impact-with-a-fixed-plane, conservation-of-momentum, direct-impact.

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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.

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
A smooth sphere of mass m=2m=2 kg slides on a smooth horizontal plane 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=5u=5 m s−1^{-1} and the coefficient of restitution between the sphere and the wall is e=35e=\frac{3}{5}. Find the velocity of the sphere immediately after the impact with the wall.

Worked solution

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

    m=2,u=5,e=35m=2,\quad u=5,\quad e=\frac{3}{5}

    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=−3v=-3

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

Answer
v=−3 m s−1v=-3\text{ m s}^{-1}
Question 3
2 markseasy
A smooth sphere of mass m=14m=\frac{1}{4} kg slides on a smooth horizontal floor 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=8u=8 m s−1^{-1} and the coefficient of restitution between the sphere and the wall is e=12e=\frac{1}{2}. Find the velocity of the sphere immediately after the impact with the wall.

Worked solution

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

    m=14,u=8,e=12m=\frac{1}{4},\quad u=8,\quad e=\frac{1}{2}

    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 4
2 markseasy
A smooth sphere of mass m=3m=3 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=10u=10 m s−1^{-1} and the coefficient of restitution between the sphere and the wall is e=14e=\frac{1}{4}. Find the velocity of the sphere immediately after the impact with the wall.

Worked solution

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

    m=3,u=10,e=14m=3,\quad u=10,\quad e=\frac{1}{4}

    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∣=14×10=52\left|v\right|=\frac{1}{4}\times10=\frac{5}{2}

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

  4. State the velocity of the sphere after the impact

    v=−52v=-\frac{5}{2}

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

Answer
v=−52 m s−1v=-\frac{5}{2}\text{ m s}^{-1}
Question 5
2 markseasy
A smooth sphere of mass m=32m=\frac{3}{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=4u=4 m s−1^{-1} and the coefficient of restitution between the sphere and the wall is e=34e=\frac{3}{4}. Find the velocity of the sphere immediately after the impact with the wall.

Worked solution

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

    m=32,u=4,e=34m=\frac{3}{2},\quad u=4,\quad e=\frac{3}{4}

    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∣=34×4=3\left|v\right|=\frac{3}{4}\times4=3

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

  4. State the velocity of the sphere after the impact

    v=−3v=-3

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

Answer
v=−3 m s−1v=-3\text{ m s}^{-1}

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