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(D) $1$. Velocity of approach: The components of velocity along the line of impact (the horizontal axis) are $v \sin(	heta / 2)$ for both particles t

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All of the above.

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(D) $1$. Velocity of approach: The components of velocity along the line of impact (the horizontal axis) are $v \sin(	heta / 2)$ for both particles towards each other. Thus,the velocity of approach is $v \sin(	heta / 2) + v \sin(	heta / 2) = 2v \sin(	heta / 2)$.$2$. Velocity of separation: In a perfectly inelastic collision,the particles stick together after the collision,so their relative velocity of separation is $0$.$3$. Common velocity after collision: By the law of conservation of linear momentum along the horizontal axis,let the mass of each particle be $m$. The initial momentum components along the horizontal axis are $mv \cos(	heta / 2)$ and $mv \cos(	heta / 2)$. After the collision,the combined mass is $2m$ moving with velocity $v'$.$mv \cos(	heta / 2) + mv \cos(	heta / 2) = (2m)v'$$2mv \cos(	heta / 2) = 2mv'$$v' = v \cos(	heta / 2)$.Since all statements are correct,the correct option is $D$.

Two identical particles are moving with same velocity $v$ as shown in the figure. If the collision is completely inelastic,then:

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