QJEEM18Ph9fd4: JEE Main 2018 Physics - Class 11 JEEM18Ph9fd4 Solution: A proton of mass m collides elastically with a particle of…

In figure (i) before collision, m' is mass of unknown particle; m is mass of proton; v₁ is initial velocity.

(i) Before collision :

Now, in figure (ii), v₁ is final velocity of unknown particle and v₂ is final velocity of proton.

(ii) After collision :

By conservation of momentum, we have

Momentum before collision = Momentum after collision

Consider x-component, we have

$m{v_i} + m'\,.\,0 = m'{v_1}\cos 45^\circ + m{v_2}\cos 45^\circ$

$m{v_i} = {1 \over {\sqrt 2 }}(m'{v_1} + m{v_2})$ ..... (1)

Consider y-component, we have

$0 = m'{v_1}\sin 45^\circ - m{v_2}\sin 45^\circ$

${1 \over {\sqrt 2 }}(m'{v_i} - m{v_2}) = 0 \Rightarrow m'{v_1} = m{v_2}$ ...... (2)

Substitute Eq. (2) in Eq. (1), we get

$m{v_i} = {1 \over {\sqrt 2 }}(m{v_2} + m{v_2}) - \sqrt 2 m{v_2}$

$\Rightarrow {v_i} = \sqrt 2 {v_2}$ ..... (3)

Using Eq. (2) and (3) in Eq. (1), we get

$m\sqrt 2 {v_2} = {1 \over {\sqrt 2 }}(m'{v_1} + m'{v_1})$

$2m{v_2} = 2m'{v_1}$ ( $\because$ ${v_1} = {v_2}$ )

$\Rightarrow m = m'$