A particle of mass m, moving with veloci...

A particle of mass `m`, moving with velocity `v` collides a stationary particle of mass `2m`. As a result of collision, the particle of mass `m` deviates by `45^@` and has final speed of `v/2`. For this situation mark out the correct statement (s).

Text Solution

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The correct Answer is:

B, D

Momentum remains conserved in any type of equation

`p_i=p_f`
`:. (mvhati)=((mv)/(2sqrt2)hati+(mv)/(2sqrt2)hatj)+(2m)v`

v=velocity of mass `2m`
`=0.32hati-0.35hatj`
`=(0.32v)hati-(0.18v)hatj`
`V=0.37v`
`tan theta=(0.18v)/(0.32v)=0.5625`
`:. theta=29.35^@`
Since `(theta+45^@)lt90^@`. Therefore, the angle of divergence between particles after collision is less than `90^@`.
Further, `K_i=1/2mv^2` and
`K_f=1/2m(v/2)^2+1/2(2m)(0.37v)^2`
`K_(f)ltK_(i)`
Therefore, collision is inelastic.

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A particle of mass m moving with velocity v strikes a stationary particle of mass 2 m and sticks to it. The speed of the system will be.

v//2

2 v

v//3

3 v

A particle of mass 2m moving with velocity v strikes a stationary particle of mass 3m and sticks to it . The speed of the system will be

`0.8 v`

`0.2 v`

`0.6v`

`0.4v`

A particle of mass m_0 , travelling at speed v_0 . Strikes a stationary particle of mass 2m_0 . As a result of the particle of mass m_0 is deflected through 45^@ and has a final speed of v_0/sqrt2 . Then the speed of the particle of mass 2m_0 after this collision is

(a) `v_0/2`

(b) `(v_0)/(2sqrt2)`

(d) `(v_0)/(sqrt2)`