3:5
5:4
4:5
To determine the ratio of the magnitudes of linear momentum for two bodies with equal kinetic energies, we need to understand the relationship between kinetic energy, mass, and momentum.
1. Write the expression for kinetic energy for each body:
2. Since the kinetic energies are equal, we equate the expressions:
\(\frac{1}{2} m_1 v_1^2 = \frac{1}{2} m_2 v_2^2\)
Cancel the \(\frac{1}{2}\) factor and rearrange the equation:
\(m_1 v_1^2 = m_2 v_2^2\)
3. From this, express the velocities in terms of each other:
\(v_1^2 = \frac{m_2}{m_1} v_2^2\)
4. Solve for the ratio of velocities:
\(\frac{v_1}{v_2} = \sqrt{\frac{m_2}{m_1}} = \sqrt{\frac{25}{4}} = \frac{5}{2}\)
5. The ratio of the magnitudes of linear momentum \((p_1/p_2)\) can be expressed as:
\(\frac{p_1}{p_2} = \frac{m_1 v_1}{m_2 v_2}\)
Substitute the values and expression for \(\frac{v_1}{v_2}\):
\(\frac{p_1}{p_2} = \frac{4 \cdot \frac{5}{2}v_2}{25 \cdot v_2} = \frac{20}{50} = \frac{2}{5}\)
The ratio of their linear momentum is \(2:5\).
The correct answer is 2:5. This matches the given correct answer in the options.
For objects with equal kinetic energies \(\left(\frac{p_1^2}{2m_1} = \frac{p_2^2}{2m_2}\right)\), we have:
\[\frac{p_1}{p_2} = \sqrt{\frac{m_1}{m_2}}\]
Substituting \(m_1 = 4 \, \text{g}\) and \(m_2 = 25 \, \text{g}\):
\[\frac{p_1}{p_2} = \sqrt{\frac{4}{25}} = \frac{2}{5}\]
Thus, the ratio of their momenta is 2 : 5.
A bob of mass \(m\) is suspended at a point \(O\) by a light string of length \(l\) and left to perform vertical motion (circular) as shown in the figure. Initially, by applying horizontal velocity \(v_0\) at the point ‘A’, the string becomes slack when the bob reaches at the point ‘D’. The ratio of the kinetic energy of the bob at the points B and C is: 
Two light beams fall on a transparent material block at point 1 and 2 with angle \( \theta_1 \) and \( \theta_2 \), respectively, as shown in the figure. After refraction, the beams intersect at point 3 which is exactly on the interface at the other end of the block. Given: the distance between 1 and 2, \( d = 4/3 \) cm and \( \theta_1 = \theta_2 = \cos^{-1} \frac{n_2}{2n_1} \), where \( n_2 \) is the refractive index of the block and \( n_1 \) is the refractive index of the outside medium, then the thickness of the block is cm. 
Two light beams fall on a transparent material block at point 1 and 2 with angle \( \theta_1 \) and \( \theta_2 \), respectively, as shown in the figure. After refraction, the beams intersect at point 3 which is exactly on the interface at the other end of the block. Given: the distance between 1 and 2, \( d = \frac{4}{3} \) cm and \( \theta_1 = \theta_2 = \cos^{-1} \left( \frac{n_2}{2n_1} \right) \), where \( n_2 \) is the refractive index of the block and \( n_1 \) is the refractive index of the outside medium, then the thickness of the block is …….. cm.

Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to:
Kinetic energy of an object is the measure of the work it does as a result of its motion. Kinetic energy is the type of energy that an object or particle has as a result of its movement. When an object is subjected to a net force, it accelerates and gains kinetic energy as a result. Kinetic energy is a property of a moving object or particle defined by both its mass and its velocity. Any combination of motions is possible, including translation (moving along a route from one spot to another), rotation around an axis, vibration, and any combination of motions.
