\( \left[ u^2 - \frac{19Gm}{10R} \right]^{\frac{1}{2}} \)
Step 1: Apply Energy Conservation The total mechanical energy at a distance \( 20R \) from the planet’s center is: \[ E_{\text{initial}} = \frac{1}{2} m u^2 - \frac{GMm}{20R} \] At the surface (\( R \)), the energy is: \[ E_{\text{final}} = \frac{1}{2} m v^2 - \frac{GMm}{R} \] Applying the conservation of energy: \[ \frac{1}{2} m u^2 - \frac{GMm}{20R} = \frac{1}{2} m v^2 - \frac{GMm}{R} \]
Step 2: Solve for \( v \) Rearranging: \[ \frac{1}{2} m v^2 = \frac{1}{2} m u^2 + \frac{GMm}{R} - \frac{GMm}{20R} \] \[ \frac{1}{2} m v^2 = \frac{1}{2} m u^2 + \frac{19GMm}{20R} \] Dividing by \( m \) and multiplying by 2: \[ v^2 = u^2 + \frac{38GM}{20R} \] \[ v^2 = u^2 + \frac{19GM}{10R} \] Taking the square root: \[ v = \left[ u^2 + \frac{19GM}{10R} \right]^{\frac{1}{2}} \] Thus, the velocity with which the object hits the surface of the planet is: \[ \mathbf{\left[ u^2 + \frac{19GM}{10R} \right]^{\frac{1}{2}}} \]
The driver sitting inside a parked car is watching vehicles approaching from behind with the help of his side view mirror, which is a convex mirror with radius of curvature \( R = 2 \, \text{m} \). Another car approaches him from behind with a uniform speed of 90 km/hr. When the car is at a distance of 24 m from him, the magnitude of the acceleration of the image of the side view mirror is \( a \). The value of \( 100a \) is _____________ m/s\(^2\).
When the object and the screen are 90 cm apart, it is observed that a clear image is formed on the screen when a convex lens is placed at two positions separated by 30 cm between the object and the screen. The focal length of the lens is:
A constant force of \[ \mathbf{F} = (8\hat{i} - 2\hat{j} + 6\hat{k}) \text{ N} \] acts on a body of mass 2 kg, displacing it from \[ \mathbf{r_1} = (2\hat{i} + 3\hat{j} - 4\hat{k}) \text{ m to } \mathbf{r_2} = (4\hat{i} - 3\hat{j} + 6\hat{k}) \text{ m}. \] The work done in the process is:
A ball 'A' of mass 1.2 kg moving with a velocity of 8.4 m/s makes a one-dimensional elastic collision with a ball 'B' of mass 3.6 kg at rest. The percentage of kinetic energy transferred by ball 'A' to ball 'B' is:
A metre scale is balanced on a knife edge at its centre. When two coins, each of mass 9 g, are kept one above the other at the 10 cm mark, the scale is found to be balanced at 35 cm. The mass of the metre scale is:
A body of mass \( m \) and radius \( r \) rolling horizontally with velocity \( V \), rolls up an inclined plane to a vertical height \( \frac{V^2}{g} \). The body is: