Question

A particle A of mass 1 g moving with a speed 2 cm/s collides with an identical particle B at rest. If coefficient of restitution is 0.5 then the ratio of velocity of A and B will be​

Answer 1

$\sf\large\bold{\underline{\pink{Given :-}}}$

Mass of Particle A $\sf{(m_{1}) = 1\:g}$

Mass of Particle B $\sf{(m_{2}) = 1\:g}$

Initial Velocity of Particle A $\sf{(u_{1}) = 2\:cm/s}$

Initial Velocity of Particle B $\sf{(u_{2}) = 0}$

Since Particles are identical that's why $\sf{m_{1} = m_{2}}$

$\sf\large\bold{\underline{\pink{To\: Find :-}}}$

$\sf{\dfrac{Final\: Velocity\:of\:A}{Final\: Velocity\:of\:B} = \dfrac{v_{1}}{v_{2}}}$

$\sf\large\bold{\underline{\pink{Solution :-}}}$

$\sf\bold{\underline{\blue{Using\: Momentum\: Conservation}}}$

$\sf\bold\green{m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}}$

$\sf{1×2 + 1×0 = 1×v_{1} + 1×v_{2}}$

$\sf{2 + 0 = v_{1} + v_{2}}$

$\sf\bold\orange{v_{1} + v_{2} = 2}$ •••• (i)

$\sf\bold{\underline{\pink{Now,}}}$

$\sf{Coefficient\: of\: Restitution = 0.5 = \dfrac{1}{2}}$

$\sf{e = \dfrac{v_{2} - v_{1}}{u_{1} - u_{2}}}$

$\sf{\dfrac{1}{2} = \dfrac{v_{2} - v_{1}}{2 - 0}}$

$\sf\bold\orange{v_{1} - v_{2} = 1}$ •••• (ii)

Solving equation (i) and (ii)

$\sf\bold\green{v_{1} = 0.5}$

$\sf\bold\green{v_{2} = 1.5}$

$\sf\bold\blue{Ratio = \dfrac{v_{1}}{v_{2}} = \dfrac{0.5}{1.5} = \dfrac{1}{3}}$

Answer 2

Answer:

The ratio of final velocity of A and B after collision is equal to 1/3.

Explanation:

Given:

The mass of body A, $m_{1}=1g$

The mass of body B, $m_{2}=1g$

Before the collision:

The velocity of body A, $u_{1}=2cms^{-1}$

The velocity of body B, $u_{2}=0$

After the collision:

The velocity of body A  and  body B are  $v_{1}$  and $v_{2}$ respectively.

From the law of conservation of momentum:

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

$(1)(2)+(1)(0)=(1)v_{1}+(1)v_{2}$

$v_{1}+v_{2}=2$                                                                  ...................(1)

The restitution coefficient = velocity of separation/ velocity of approach

$e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}}$

$0.5=\frac{v_{2}-v_{1}}{2-0}$

$v_{2}-v_{1}=1$

$v_{2}=1+v_{1$

Put the value of v₁ in eq.(1);

$v_{1}+1+v_{1}=2\\2v_{1}=1$

$v_{1}=\frac{1}{2}$

and, $v_{2}=1+\frac{1}{2}$

        $v_{2}=\frac{3}{2}$                                                

The ratio of   $\frac{v_{1}}{v_{2}} =\frac{1}{2}*\frac{2}{3}$

Therefore,    $\frac{v_{1}}{v_{2}}=\frac{1}{3}$