Elastic collisions in two dimensions Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Elastic collisions in two dimensions questions. See exactly how to solve problems on oblique-impact, impact-with-a-fixed-plane, restitution, kinetic-energy-loss.

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.

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=2m=2 kg is sliding on a smooth horizontal plane 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=6u_{t}=6 m s−1^{-1} (parallel to the wall) and un=8u_{n}=8 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=23e=\frac{2}{3}. Find the components of its velocity immediately after the impact with the wall.

Worked solution

  1. Write down the data for the impact

    m=2,un=8,ut=6,e=23m=2,\quad u_{n}=8,\quad u_{t}=6,\quad e=\frac{2}{3}

    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=6v_{t}=u_{t}=6

    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=(−163, 6)\mathbf{v}=\left(-\frac{16}{3},\ 6\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=(−163, 6) m s−1\mathbf{v}=\left(-\frac{16}{3},\ 6\right)\text{ m s}^{-1}
Question 3
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=12u_{n}=12 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=35e=\frac{3}{5}. Find the components of its velocity immediately after the impact with the wall.

Worked solution

  1. Write down the data for the impact

    m=1,un=12,ut=5,e=35m=1,\quad u_{n}=12,\quad u_{t}=5,\quad e=\frac{3}{5}

    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=5v_{t}=u_{t}=5

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

  3. Reverse and scale the normal component

    vn=−e un=−35×12=−365v_{n}=-e\,u_{n}=-\frac{3}{5}\times12=-\frac{36}{5}

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

  4. State the velocity immediately after the impact

    v=(−365, 5)\mathbf{v}=\left(-\frac{36}{5},\ 5\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=(−365, 5) m s−1\mathbf{v}=\left(-\frac{36}{5},\ 5\right)\text{ m s}^{-1}
Question 4
2 markseasy
A smooth sphere of mass m=3m=3 kg is sliding on a smooth ice rink 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=14e=\frac{1}{4}. Which of the following gives the components of its velocity immediately after the impact with the wall?

Worked solution

  1. Write down the data for the impact

    m=3,un=6,ut=8,e=14m=3,\quad u_{n}=6,\quad u_{t}=8,\quad e=\frac{1}{4}

    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=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.

  3. Reverse and scale the normal component

    vn=−e un=−14×6=−32v_{n}=-e\,u_{n}=-\frac{1}{4}\times6=-\frac{3}{2}

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

  4. State the velocity immediately after the impact

    v=(−32, 8)\mathbf{v}=\left(-\frac{3}{2},\ 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.

Answer
v=(−32, 8) m s−1\mathbf{v}=\left(-\frac{3}{2},\ 8\right)\text{ m s}^{-1}
Question 5
2 markseasy
A smooth sphere of mass m=14m=\frac{1}{4} kg is sliding on a smooth horizontal surface 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=3u_{t}=3 m s−1^{-1} (parallel to the wall) and un=4u_{n}=4 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=13e=\frac{1}{3}. Which of the following gives the components of its velocity immediately after the impact with the wall?

Worked solution

  1. Write down the data for the impact

    m=14,un=4,ut=3,e=13m=\frac{1}{4},\quad u_{n}=4,\quad u_{t}=3,\quad e=\frac{1}{3}

    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=3v_{t}=u_{t}=3

    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=(−43, 3)\mathbf{v}=\left(-\frac{4}{3},\ 3\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=(−43, 3) m s−1\mathbf{v}=\left(-\frac{4}{3},\ 3\right)\text{ m s}^{-1}

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