The bob of a pendulum of length $l$ is pulled aside from its equilibrium position through an angle $\theta $ and then released. The bob will then pass through its equilibrium position with speed $v$ , where $v$ equals
$\sqrt {2gl\,\sin \,\theta } $
$\sqrt {2gl\,\left( {1 - \sin \,\theta } \right)} $
$\sqrt {2gl\,\left( {1 - \cos \,\theta } \right)} $
$\sqrt {2gl\,\left( {1 + \sin \,\theta } \right)} $
A mass $m$ moving horizontally with velocity $v_0$ strikes a pendulum of mass $m$. If the two masses stick together after the collision, then the maximum height reached by the pendulum is
The potential energy of a body of mass $m$ is:
$U = ax + by$
Where $x$ and $y$ are position co-ordinates of the particle. The acceleration of the particle is
The kinetic energy acquired by a body of mass m is travelling some distance s, starting from rest under the actions of a constant force, is directly proportional to
$A$ man who is running has half the kinetic energy of the boy of half his mass. The man speeds up by $1 \, m/s$ and then has the same kinetic energy as the boy. The original speed of the man was
In an elastic collision of two particles the following quantity is conserved