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Relation between kinetic energy and mome...

Relation between kinetic energy and momentum
Let us consider a body of mass 'm' having a velocity 'v', then
momentum of the body P = mass `xx` velocity
`P = m xx v rArr v = (P)/(m) " "...(1)`
From definition, kinetic energy (K.E) of the body
`K.E = (1)/(2) mv^(2)" "...(2)`
Now putting the value of (1) in (2) we have
`K.E = (1)/(2)m ((P)/(m))^(2)`
`K.E. = (1)/(2)m (P^(2))/(m^(2)) = (1)/(2) (P^(2))/(m) = (P^(2))/(2m)" "...(3)`
Thus we can write
`P^(2) = 2m xx K.E`
`rArr P = sqrt(2m xx K.E)`
Thus momentum = `sqrt(2 xx "mass" xx "kinetic energy")`
The kinetic energy of a given body is doubled. Its momentum will

A

remain unchanged

B

redoubled

C

become `(1)/(2)` times

D

become `sqrt(2)` times

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Relation between kinetic energy and momentum Let us consider a body of mass 'm' having a velocity 'v', then momentum of the body P = mass xx velocity P = m xx v rArr v = (P)/(m) " "...(1) From definition, kinetic energy (K.E) of the body K.E = (1)/(2) mv^(2)" "...(2) Now putting the value of (1) in (2) we have K.E = (1)/(2)m ((P)/(m))^(2) K.E. = (1)/(2)m (P^(2))/(m^(2)) = (1)/(2) (P^(2))/(m) = (P^(2))/(2m)" "...(3) Thus we can write P^(2) = 2m xx K.E rArr P = sqrt(2m xx K.E) Thus momentum = sqrt(2 xx "mass" xx "kinetic energy") What will be the momentum of a body of mass 100 g having kinetic energy of 20 J ?

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Knowledge Check

  • Relation between Kinetic energy (Ex) and Momentum (p) of a body of mass (m)

    A
    `E_k = (p^2)/(m)`
    B
    `E_(k) = (p^2)/(3m)`
    C
    `E_k = (p)/(2m)`,
    D
    `E_k = (p^2)/(2m)`

  • The kinetic energy of a body of mass 2 kg and momentum of 2 Ns is

    A
    1J
    B
    2J
    C
    3J
    D
    4J

  • The kinetic energy of a body of mass 2 kg and momentum of 2 Ns is

    A
    1 J
    B
    2 J
    C
    3 J
    D
    4 J

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    Relation between kinetic energy and momentum Let us consider a body of mass 'm' having a velocity 'v', then momentum of the body P = mass xx velocity P = m xx v rArr v = (P)/(m) " "...(1) From definition, kinetic energy (K.E) of the body K.E = (1)/(2) mv^(2)" "...(2) Now putting the value of (1) in (2) we have K.E = (1)/(2)m ((P)/(m))^(2) K.E. = (1)/(2)m (P^(2))/(m^(2)) = (1)/(2) (P^(2))/(m) = (P^(2))/(2m)" "...(3) Thus we can write P^(2) = 2m xx K.E rArr P = sqrt(2m xx K.E) Thus momentum = sqrt(2 xx "mass" xx "kinetic energy") Two bodies of mass 1 kg and 4 kg possess equal momentum. The ratio of their kinetic energies is

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