A body of mass $$10 \mathrm{~kg}$$ is projected at an angle of $$45^{\circ}$$ with the horizontal. The trajectory of the body is observed to pass through a point $$(20,10)$$. If $$\mathrm{T}$$ is the time of flight, then its momentum vector, at time $$\mathrm{t}=\frac{\mathrm{T}}{\sqrt{2}}$$, is _____________.

[Take $$\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$$ ]

A block of mass M slides down on a rough inclined plane with constant velocity. The angle made by the incline plane with horizontal is $$\theta$$. The magnitude of the contact force will be :

A block 'A' takes 2 s to slide down a frictionless incline of 30$$^\circ$$ and length 'l', kept inside a lift going up with uniform velocity 'v'. If the incline is changed to 45$$^\circ$$, the time taken by the block, to slide down the incline, will be approximately :

The velocity of the bullet becomes one third after it penetrates 4 cm in a wooden block. Assuming that bullet is facing a constant resistance during its motion in the block. The bullet stops completely after travelling at (4 + x) cm inside the block. The value of x is :