Question 203: Centre of Mass and Rotational Motion

Question

A proton of mass m collides elastically with a particle of unknown mass at rest. After the collision,  the proton and the unknown particle are seen moving at an angle of 90
o
with respect to each other. The mass of unknown particle is :

Accepted Answer

In figure (i) before collision, m' is mass of unknown particle; m is mass of proton; v
1
is initial velocity.
(i) Before collision :
Now, in figure (ii), v
1
is final velocity of unknown particle and v
2
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
$$m\sqrt 2 {v_2} = {1 \over {\sqrt 2 }}(m'{v_1} + m'{v_1})$$
$$2m{v_2} = 2m'{v_1}$$ ($$\because$$ $${v_1} = {v_2}$$)
$$ \Rightarrow m = m'$$

Answer

$${m \over 2}$$

Answer

m

Answer

$${m \over {\sqrt 3 }}$$

Answer

2 m

Centre of Mass and Rotational Motion

A proton of mass m collides elastically with a particle of unknown mass at rest. After the collision, the proton and the unknown particle are seen moving at an angle of 90 o with respect to each other. The mass of unknown particle is :

Choose the correct answer: