Question 201: 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 90o 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 90o with respect to each other. The mass of unknown particle is :

Choose the correct answer: