A spring is compressed between two toy carts to masses `m_1` and `m_2`. When the toy carts are released, the spring exerts on each toy cart equal and opposite forces for the same small time t. If the coefficients of friction `mu` between the ground and the toy carts are equal, then the magnitude of displacements of the toy carts are in the ratio

Options

A. (a) `S_1/S_2=m_2/m_1`
B. (b) `S_1/S_2=m_1/m_2`
C. (c) `S_1/S_2=(m_2/m_1)^2`
D. (d) `S_1/S_2=(m_1/m_2)^2`

Answer

C

Explanation

Minimum stopping distance=s
Work done against the friction`=W=mumgs`
Initial momentum gained by both toy carts will be same because same force acts for same time.
Initial kinetic energy of the toy cart `=((p^2)/(2m))`
Therefore, `mumgs=(p^2)/(2m)` or `s=((p^2)/(2mugm^2))`
For the two toy carts, momentum is numerically the same.
Further `mu` and g are the same for the toy carts.
So, `s_1/s_2=(m_2/m_1)^2`