A bullet of mass \(10^{-2}\) kg and velocity \(200\) m/s gets embedded inside the bob of mass \(1\) kg of a simple pendulum. The maximum height that the system rises by is_____ cm.
The Correct Answer is : 20
Momentum conservation :
\(10^{-2} \times 200 ≃ 1 × v …(1)\)
Energy conservation :
v\(=\sqrt{2gh}\ \ \ ....(2)\)
\(⇒h=\frac{v^2}{2g}\)
\(=\frac{4}{20}m=20cm\)
From a height of 'h' above the ground, a ball is projected up at an angle \( 30^\circ \) with the horizontal. If the ball strikes the ground with a speed of 1.25 times its initial speed of \( 40 \ ms^{-1} \), the value of 'h' is:
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $
Resonance in X$_2$Y can be represented as
The enthalpy of formation of X$_2$Y is 80 kJ mol$^{-1}$, and the magnitude of resonance energy of X$_2$Y is:
We can note there involves a continuous interchange of potential and kinetic energy in a simple harmonic motion. The system that performs simple harmonic motion is called the harmonic oscillator.
Case 1: When the potential energy is zero, and the kinetic energy is a maximum at the equilibrium point where maximum displacement takes place.
Case 2: When the potential energy is maximum, and the kinetic energy is zero, at a maximum displacement point from the equilibrium point.
Case 3: The motion of the oscillating body has different values of potential and kinetic energy at other points.