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