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

Options

A. `(m_(1))/(m_(2))`
B. `(m_(2))/(m_1)`
C. 1
D. `(sqrt(m_(1) m_(2)))/(m_(1) + m_(2))`

Answer

B

Explanation

The acceleration of the body of mass `m_(1)` acts upon by a constant F is
`a_(1) = (F)/(m_1) " " … (i)`
The acceleration of the body of mass `m_(2)` acts upon by a same constant F is
`a_(2) = (F)/(m_(2)) " " … (ii)`
Starting from rest , the velocity acquired by mass `m_(1)` in time t is
`v_(1) = a_(1) t = (F)/(m_(1)) "" t` (Using (i)) `" " ... (iii)`
Starting from rest , the velocity acquired by mass `m_(2)` in the same time t is
`v_(2) = a_(2) t = (F)/(m_(2)) t ` (Using (ii)) `" " ... (iv)`
Divide (iii) by (iv) , we get
`therefore (v_(1))/(v_(2)) = (m_(2))/(m_(1)) " " ... (v)`
Kinetic energy of mass `m_(1)` is
`E_(1) = (1)/(2) m_(1) v_(1)^(2) " " .... (vi)`
Kinetic energy of mass `m_(2)` is
`E_(2) = (1)/(2) m_(2) v_(2)^(2) " " ... (vii)`
Divide (vi) by (vii) we get
`(E_(1))/(E_(2)) = (m_(1))/(m_(2)) ((v_(1))/(v_(2)))^(2) = (m_(1))/(m_(2)) ((m_(2))/(m_(1)))^(2) " " ` (Using (v))
`= (m_(2))/(m_(1))`