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Two bodies of masses m1 and m2 are acted...

Two bodies of masses `m_1` and `m_2` are acted upon by a constant force F for a time t . They start from rest and acquire kinetic energies `E_(1)` and `E_(2)` respectively . Then `(E_(1))/(E_(2))` is

A

`(m_1)/(m_2)`

B

`(m_2)/(m_1)`

C

1

D

`sqrt((m_1 m_2)/(m_1+m_2))`

Text Solution

Verified by Experts

The correct Answer is:
b
acceleration of mass `m_1` is `a_1=(F)/(m_1)`
`:. V_1=u+a_1t=0+(F)/(m_1).t`
`:. E_1=1/2mv_1^2=1/2m_1.(F^2)/(m_1^2).t=(F^2t)/(2m_1)`
Similarly `a_2=(F)/(m_2)`
`v_2=u+a_2t=0+(F)/(m_2).t`
`:. E_2=1/2m_2v_2^2=1/2m_2.(F^2)/(m_2^2)t^2=(F)/(m_1)=(F^2t)/(m_2)`
`:.` `(E_1)/(E_2)=(m_2)/(m_1)`
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