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प्रश्न
Answer the following question.
A bullet of mass m1 travelling with a velocity u strikes a stationary wooden block of mass m2 and gets embedded into it. Determine the expression for loss in the kinetic energy of the system. Is this violating the principle of conservation of energy? If not, how can you account for this loss?
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उत्तर
- A bullet of mass m1 travelling with a velocity u, striking a stationary wooden block of mass m2 and getting embedded into it is a case of perfectly inelastic collision.
- In a perfectly inelastic collision, although there is a loss in kinetic energy, the principle of conservation of energy is not violated as the total energy of the system is conserved.
Loss in the kinetic energy during a perfectly inelastic head-on collision:
- Let two bodies A and B of masses m1 and m2 moving with initial velocity `vec"u"_1 and vec"u"_2` respectively such that particle A collides headon with particle B i.e., `"u"_1 > "u"_2`.
- If the collision is perfectly inelastic, the particles stick together and move with a common velocity `vec"v"` after the collision along the same straight line.
loss in kinetic energy = total initial
kinetic energy – total final kinetic energy, - By the law of conservation of momentum, m1u1 + m2 u2 = (m1 + m2) v
∴ v = `("m"_1"u"_1 + "m"_2"u"_2)/("m"_1 + "m"_2)` - Loss of kinetic energy,
`Delta "K.E" = (1/2"m"_1"u"_1^2 + 1/2"m"_2"u"_2^2) - 1/2("m"_1 + "m"_2)"v"^2`
`= (1/2"m"_1"u"_1^2 + 1/2"m"_2"u"_2^2) -1/2("m"_1 + "m"_2)[("m"_1"u"_1 + "m"_2"u"_2)/("m"_1 + "m"_2)]^2`
`= ("m"_1^2"u"_1^2 + "m"_1"m"_2"u"_2^2 + "m"_1"m"_2"u"_1^2)/(2("m"_1 + "m"_2)) + ("m"_2^2 "u"_2^2 - "m"_1^2"u"_1^2 - "m"_2^2"u"_2^2 - 2"m"_1"m"_2"u"_1"u"_2)/(2("m"_1 + "m"_2))`
`= ("m"_1"m"_2)/(2("m"_1 + "m"_2)) ("u"_1 - "u"_2)^2` - Both the masses and the term `("u"_1 - "u"_2)^2` are positive. Hence, there is always a loss in a perfectly inelastic collision. For a perfectly inelastic collision, as e = 0, the loss is maximum.
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