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 one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! 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.