A force $$\mathbf{F}_1$$ of magnitude 4 N acts on an object of mass 1 kg , at origin in a direction $$30^{\circ}$$ above the positive $$X$$-axis. A second $$F_2$$ of magnitude 4 N acts on the same object in the direction of the positive $$Y$$-axis. The magnitude of the acceleration of the object is nearly.
$$y=\left(P t^2-Q t^3\right) \mathrm{~m}$$ is the vertical displacement of a ball which is moving in vertical plane. Then the maximum height that the ball can reach is
A cricket ball of mass 50 g having velocity $$50 \mathrm{~cm} \mathrm{~s}^{-1}$$ to stopped in 0.5 s. The force applied to stop the ball is
Two masses $$M_1$$ and $$M_2$$ are arranged as shown in the figure. Let $$a$$ be the magnitude of the acceleration of the mass $$M_1$$. If the mass of $$M_1$$ is doubled and that of $$M_2$$ is halved, then the acceleration of the system is (Treat all surfaces as smooth; masses of pulley and rope are negligible)