A force F=VXA is applied on a particle of mass m in addition to the force of gravity, where A is a constant vector in the horizontal direction and V is velocity vector.Find the minimum speed with which the particle must be projected so that it moves undeflected with a constant velocity.

Hi,

Concept: V(cross)A is a vector perpendicular to the plane containing V(vector) and (A vector).

Consider horizontal plane as the X-Y plane, so that gravity is perpendicular to this plane (along -ve Z-direction)

And this vector froce, F should cancel out the vecotr force due to gravity.

Let's assume that gravity is in the (Z - axis direction), which would be = mg(downwards)

Now the force you apply should exactly cancel out this force. Hence F(vector) should be in the upwards direction with magnitude = mg.

Mag |F| = mg = VAsin(Q); where Q is the angle between the vector V and vector A. (Both V and A in the horizontal plane X-Y), so that their cross product F is in the vertical direction.

So |V| should be mg/(|A|*sinQ)

now for the minimum speed, sinQ should be maximum (which we know is 1), hence Q = 90 degrees.

Hence V(vector) should be perpendicular to A(vector) in the horizontal plane.

And hence if A is in the posive x-direction, V should be along the negative Y-direction, so that V(cross)A is along the positive Z-direction, to cancel out the effect of gravity along the negative Z-direction.

Hope that was clear and helpful.

All the best

Regards,

Ashwin (IIT Madras)