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    4.1 Newton's Law of Restitution For Direct Impact

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    GM
    The document describes Newton's Law of Restitution, which relates the speeds of particles before and after collision. It states that the speed of separation is equal to the coefficient of restitution (e) multiplied by the speed of approach. e is a constant between 0 and 1 that depends on the materials, with 1 being a perfectly elastic collision and 0 being completely inelastic. Diagrams show example collisions and the law is applied to calculate speeds before and after collision given the coefficient of restitution.

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    4.1 Newton's Law of Restitution For Direct Impact

    Uploaded by

    GM
    22 views5 pages
    The document describes Newton's Law of Restitution, which relates the speeds of particles before and after collision. It states that the speed of separation is equal to the coefficient of restitution (e) multiplied by the speed of approach. e is a constant between 0 and 1 that depends on the materials, with 1 being a perfectly elastic collision and 0 being completely inelastic. Diagrams show example collisions and the law is applied to calculate speeds before and after collision given the coefficient of restitution.

    Original Description:

    Original Title

    4.1 NLR v2

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    The document describes Newton's Law of Restitution, which relates the speeds of particles before and after collision. It states that the speed of separation is equal to the coefficient of restitution (e) multiplied by the speed of approach. e is a constant between 0 and 1 that depends on the materials, with 1 being a perfectly elastic collision and 0 being completely inelastic. Diagrams show example collisions and the law is applied to calculate speeds before and after collision given the coefficient of restitution.

    Copyright:

    © All Rights Reserved
    Download as DOCX, PDF, TXT or read online from Scribd
    Download as docx, pdf, or txt
    22 views5 pages

    4.1 Newton's Law of Restitution For Direct Impact

    Uploaded by

    GM
    The document describes Newton's Law of Restitution, which relates the speeds of particles before and after collision. It states that the speed of separation is equal to the coefficient of restitution (e) multiplied by the speed of approach. e is a constant between 0 and 1 that depends on the materials, with 1 being a perfectly elastic collision and 0 being completely inelastic. Diagrams show example collisions and the law is applied to calculate speeds before and after collision given the coefficient of restitution.

    Copyright:

    © All Rights Reserved
    Download as DOCX, PDF, TXT or read online from Scribd
    Download as docx, pdf, or txt
    of 5

    4.1 Newton’s Law of Restitution for Direct Impact.

     A direct impact is a collision between two particles moving along the


    same straight line.

    When two particles collide their speeds after the collision depend on the
    materials they are made from. Similarly, when an elastic particle strikes a
    surface, such as a wall, the speed of rebound depends upon the material it is
    made from.

     Newton’s Law of Restitution shows a relationship between the


    speeds of the particles, before and after collision, and the nature of
    the particles themselves.

    Newton’s Law of Restitution (NLR) states that:

    Speed of separation of particles = e


    Speed of approach of particles

    Where e is the coefficient of restitution, which is a constant.

    NB: e is the ratio of two speeds which are +ve and 0  e  1

    e=1  two perfectly elastic particles

    e=0  two inelastic particles (which coalesce on impact)

    Most particles will have a value of e somewhere between these two.

    eg: for 2 colliding glass marbles, e = 0.95

    for 2 colliding billiard balls, e = 0.8

    for 2 colliding lead spheres, e = 0.2

    NB: this law only holds when the collision is on a smooth surface or in free
    space.
    The following diagrams show three typical situations.

    (1) Before. After.

    u1 v2

    Speed of approach = u1 – 0 speed of separation = v2 – 0


    = u1 = v2

    (2) Before. After.

    u1 u2 v1 v2

    For collision to occur for separation to occur


    u1 > u 2 v2 > v1

     speed of approach = u1 – u2  speed of separation = v2 – v1

    (3) Before. After.

    u1 u2 v1 v2

    Speed of approach = u1 – (– u2) speed of separation = v2 – (– v1)


    = u 1 + u2 = v 1 + v2
    Example 4.
    Find e in the following situations.

    (a) Before. After.

    6 ms-1 0 2 ms-1 4 ms-1

    (b) Before. After.

    4 ms-1 2 ms-1 1 ms-1 3 ms-1

    Example 5.
    Find the value of v shown when the coefficient of restitution is 

    Before. After.

    6 ms-1 3 ms-1 v ms-1 7 ms-1


    Example 6.
    Two particles A and B of mass 0.2 kg and 0.5 kg respectively
    are moving towards each other along the same straight line on a smooth
    horizontal table. A has speed 12 ms-1 and B has speed 2 ms-1. Given that the
    coefficient of restitution between the two particles is , find:
    (a) the speeds of A and B after the impact
    (b) the magnitude of the impulse given to each particle.

    Describe a realistic situation that this could be used to model.

    (a) Before. After.

    12 ms-1 2 ms-1 v1 ms-1 v2 ms-1

    m1 = 0.2 m2 = 0.5 m1 = 0.2 m2 = 0.5


    Using algebra in this way, an alternative way of writing the formula for the
    coefficient of restitution is:

    Before. After.

    u1 u2 v1 v2

    e = v2 – v 1
    u 1 – u2

    EXERCISE 4B, p105

    Questions:-

     Remember - draw a labelled Before & After diagram for each question. 

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