Two blocks `m_(1)` and `m_(2)` are pulled on a smooth horizontal surface, and are joined together with a spring of stiffness `k` as shown in Fig. Suddenly, block `m_(2)` receives a horizontal velocity `v_(0)`, then the maximum extension `x_(m)` in the spring is

Options

A. `v_(0)sqrt((m_(1)m_(2))/(m_(1)+m_(2)))`
B. `v_(0)sqrt((2m_(1)m_(2))/(((m_(1)+m_(2))k))`
C. `v_(0)sqrt((m_(1)m_(2))/(2(m_(1)+m_(2))k))`
D. `v_(0)sqrt((m_(1)m_(2))/((m_(1)+m_(2))k))`

Answer

D

Explanation

Using conservation of linear momentum we have
`m_(2) v_(0)=(m_(1)+m_(2))vimpliesv=m_(2)/(m_(1)+m_(2))v_(0)`
form work energy conservation.
`1/2mv_(2)v_(0)^(2)-1/2(m_(1)+m_(2)) (m_(2)^(2)v_(0)^(2))/((m_(1)+m_(2))^(2))=1/2kx^(2)`
`x=v_(0)sqrt((m_(1)m_(2))/(k(m_(1)+m_(2))))`