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Answer the following question. Obtain its value for an elastic collision and a perfectly inelastic collision.

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Question

Answer the following question.

Obtain its value for an elastic collision and a perfectly inelastic collision.

Answer in Brief
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Solution

  1. Consider a head-on collision of two bodies of masses m1 and m2 with respective initial velocities u1 and u2. As the collision is head-on, the colliding masses are along the same line before and after the collision. The relative velocity of approach is given as,
    ua = u2 - u1
    Let v1 and v2 be their respective velocities after the collision. The relative velocity of recede (or separation) is then vs = v2 – v1
    ∴ e = `- "v"_"s"/"u"_"a" = - ("v"_2 - "v"_1)/("u"_2 - "u"_1) = ("v"_1 - "v"_2)/("u"_2 - "u"_1)`        .....(1)
  2. For a head-on elastic collision, According to the principle of conservation of linear momentum,
    Total initial momentum = Total final momentum
    ∴ m1u1 + m2u2 = m1v1 + m2v2    ...(2)
    ∴ m1(u1 - v1) = m2(v2 - u2)    ......(3)
    As the collision is elastic, the total kinetic energy of the system is also conserved.
    ∴ `1/2 "m"_1"u"_1^2 + 1/2"m"_2"u"_2^2 = 1/2 "m"_1"v"_1^2 + 1/2 "m"_2"v"_2^2`      .....(4)
    ∴ `"m"_1("u"_1^2 - "v"_1^2) = "m"_2("v"_2^2 - "u"_2^2)`
    ∴ m1(u1 + v1)(u1 - v1) = m2(v2 + u2)(v2 - u2)     .....(5)
    Dividing equation (5) by equation (3), we get
    u1 + v1 = u2 + v2
    ∴ u2 - u1 = v1 - v2      .....(6)
    Substituting this in equation (1),
    e = `("v"_1 - "v"_2)/("u"_2 - "u"_1)` = 1
  3. For a perfectly inelastic collision, the colliding bodies move jointly after the collision, i.e.,
    v1 = v2
    ∴ v1 - v2 = 0
    Substituting this in equation (1),
    e = 0
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Chapter 4: Laws of Motion - Exercises [Page 75]

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Balbharati Physics [English] Standard 11 Maharashtra State Board
Chapter 4 Laws of Motion
Exercises | Q 2. (xiv) | Page 75

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