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Let us consider a body , A, of arbitrary geometrical shape of mass m1. Now let it be translating with a speed v. Let us consider a small point particle ,P, of mass m2. Suppose it strikes A with a speed V2. Now Will the body A rotate for all the possible ways of P striking it? If yes , Then about which axis it will rotate?

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Please do not introduce variables that you never use again, I will clean up the question. – Ron Maimon Sep 17 '11 at 18:12
This depends very strongly on the shape of the object and the scattering behaviour of the particle off of the object. – Jerry Schirmer Sep 17 '11 at 18:32
@Jerry Schirmer- Could you elaboarate ? – Primeczar Sep 17 '11 at 18:38
I don't understand what you mean by "Will the body A rotate for all the possible ways of P striking it?" – David Zaslavsky♦ Sep 17 '11 at 18:45
If I may offer a little advice, you seem to be trying to ask very general questions without having obtained a clear understanding of all the concepts at play in your questions. Doing physics relies on terms with very precise meanings. At the beginning it is easier to comprehend these terms in simple, idealized situations. Yes, this means that introductory physics can seem a little divorced from the real world, but it reduces confusion. – dmckee♦ Sep 17 '11 at 19:12
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The reflecting object P transfers a certain amount of momentum to the rigid body A, which is the vector $\Delta P$, which depends on the incoming and outgoing angle. It also gives an angular momentum which is $R \times \Delta P$ where $R$ is the vector from the center of mass to the point of contact. The object will rotate unless the angular momentum transferred is zero, which happens only when the reflection momentum transfer $\Delta P$ is in the same direction of the line from the center of mass to the point of contact.

When the object A is set rotating, the bouncing point P will bounce off slower than when the object A is not set rotating. There is also the question of whether the point has internal rotation or not, and whether the reflection has friction which can change the direction of rotation. It's a moderately complicated problem, with no obvious motivation.

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What is "the reflecting object"? What is "the rigid body"? Maybe You should try not to introduce variables You never use again? – Georg Sep 17 '11 at 19:12
@Georg: fixed it. – Ron Maimon Sep 17 '11 at 19:40
@Ron Maimon- Dear Ron , could you tell me what would have been the scenario if you were considering the angular momentum about a point other than the centre of mass? Why should we always consider the centre of mass as the origin? – Primeczar Sep 18 '11 at 5:18
@Ron Maimon - Could you tell me what you mean by obvious motivations? – Primeczar Sep 18 '11 at 5:19
@Primeczar: by "no obvious motivation" I mean that I don't know why you care about this. The reason you use the center of mass to determine how the object rotates is because there is a theorem that the angular momentum of a moving rigid body is determined by the torque around the center of mass, even when the object is accelerating. You can calculate the angular momentum around any point, but that point has to be unaccelerated for the conservation law to work. The center of mass is an exception, it works for rigid bodies always. – Ron Maimon Sep 18 '11 at 5:23
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If the relative velocity at the point of impact is $\Delta \vec{v} = \vec{v_2}-\vec{v_{1P}} $, the direction of travel for P is $\vec{n}$, and the distance from the center of gravity to the impact point for body A is $\vec{r}$ then the impulse magnitude is

$$ J = \frac{ (\epsilon+1) \vec{n}\cdot\Delta\vec{v}} {\frac{1}{m_1} + \frac{1}{m_2} - \vec{n}\cdot \frac{ \vec{r}\times\vec{r}\times\vec{n} }{I_1} } $$

where $\epsilon$ is the coefficient of restituion, and $I_1$ is the mass moment of inertia for body A. If you find the final velocity of the center of gravity $\vec{v_1}$ and rotational $\vec\omega_1$ then the instant center of rotation is located at (relative to the CG)

$$ \vec r_{\rm pole} = - \frac{ \vec\omega_1 \times \vec v_1 } { \vec \omega_1 \cdot \vec \omega_1 } $$

Note the [$\cdot$] operator is the dot product and [$\times$] the cross product.

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