G L Mittal and TARUN MITTAL Solutions for Chapter: Motion in One and Two Dimensions, Exercise 3: FOR DIFFERENT COMPETITIVE EXAMINATIONS

An object, moving with a speed of $6.25 {\mathrm{ms}}^{-1}$, is decelerated at a rate given by$\frac{dv}{dt}=-2.5\sqrt{v}$, where $v$ is the instantaneous speed. The time taken by the object, to come to rest, would be:

A particle is moving with velocity $\overrightarrow{v}=k(y\hat{i}+x\hat{j})$, where $k$ is a constant. The general equation for its path is:

A projectile is fired from the surface of the earth with a velocity of $5 m{s}^{-1}$ and angle $\theta$ with the horizontal. Another projectile fired from another planet with a velocity of $3 m{s}^{-1}$ at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is (in $m{s}^{-2}$)

Two fixed frictionless inclined planes making an angle $30°$ and $60°$ with the vertical are shown in the figure. Two blocks $A$ and $B$ are placed on the two planes. What is the relative vertical acceleration of $A$ with respect to $B$.

A particle of mass $m$ is projected from the ground with an initial speed $u_0$ at an angle $\alpha$ with the horizontal. At the highest point of its trajectory, it makes a completely inelastic collision with another identical particle, which was thrown vertically upward from the ground with the same initial speed $u_0$. The angle that the composite system makes with the horizontal immediately after the collision is:

A particle of mass $m$ is projected with velocity $v$ making an angle of $45°$ with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be

A particle is projected at $60°$ to the horizontal with a kinetic energy $K$. The kinetic energy at the highest point is

A projectile is given an initial velocity of $(\hat{i}+2\hat{j}) m{s}^{-1}$, when $\hat{i}$ is along the ground and $\hat{j}$ is along the vertical. If $g=10 m{s}^{-2}$, the equation of its trajectory is: