A spring is compressed between two blocks of masses m1 and m2 placed on a horizontal frictionless surface as shown in the figure. When the blocks are released, they have initial velocity of v1 and v2 as shown. The blocks travel distances x1 and x2 respectively before coming to rest. The ratio ((x1/x2)) is

Question

A spring is compressed between two blocks of masses m1 and m2 placed on a horizontal frictionless surface as shown in the figure. When the blocks are released, they have initial velocity of v1 and v2 as shown. The blocks travel distances x1 and x2 respectively before coming to rest. The ratio ((x1/x2)) is (A) (m2/m1) (B) (m1/m2) (C) √(m2/m1) (D) √(m1/m2)

Tardigrade Admin · Accepted Answer

Initial momentum of the system is zero i.e.

P_{i} = 0

Let the velocity acquire by masses

m_{1}

and

m_{2}

just after they are released be

v_{1}

and

v_{2}

Final momentum of the system

P_{f} = m_{1} v_{1} - m_{2} v_{2}

Using conservation of momentum :

P_{i} = P_{f}

∴ 0 = m_{1} v_{1} - m_{2} v_{2}

⟹ m_{1} v_{1} = m_{2} v_{2} (v = \frac{d x}{d t}) ⟹ m_{1} x_{1} = m_{2} x_{2} ⟹ \frac{x _{1}}{x _{2}} = \frac{m _{2}}{m _{1}}

$\frac{m _{2}}{m _{1}}$

$\frac{m _{1}}{m _{2}}$

$\frac{m _{2}}{m _{1}}$

$\frac{m _{1}}{m _{2}}$

m1​m2​​

m2​m1​​

m1​m2​​​

m2​m1​​​

$\frac{m _{2}}{m _{1}}$

$\frac{m _{1}}{m _{2}}$

$\frac{m _{2}}{m _{1}}$

$\frac{m _{1}}{m _{2}}$