The bob of simple pendulum (mass \(\mathrm{m}\) and length 1 ) dropped from a horizontal position strike a block of the same mass elastically placed on a horizontal frictionless table. The K.E. of the block will be (A) \(2 \mathrm{mg} 1\) (B) \(\mathrm{mg} 1 / 2\) (C) \(\mathrm{mg} 1\) (D) zero

Initially, the pendulum is in a horizontal position with only potential energy due to its height and the block is at rest on the frictionless table. After collision, the pendulum and the block have kinetic energy from the transferred potential energy of the pendulum.

When the pendulum is horizontally positioned, the entire length of the pendulum (1) is lifted against gravity. The potential energy (PE) at this point can be calculated using the formula: \(PE = mgh\) where m = mass of the bob g = acceleration due to gravity h = height of the bob In this case, h equals the entire length of the pendulum, which is 1. Therefore, \(PE = mg * 1 = mg\)

According to the principle of energy conservation, the total mechanical energy before collision equals the total mechanical energy after collision. Initially, the pendulum has potential energy (PE) and the block has a kinetic energy of zero. After the collision, the pendulum has some kinetic energy (KE1) and the block has some kinetic energy (KE2). To find the total mechanical energy after collision, we can equate the initial mechanical energy (PE) with the final mechanical energy (KE1 + KE2): \(PE = KE1 + KE2\) Since we are interested in finding the kinetic energy of the block (KE2), we can rewrite this equation as: \(mg = KE1 + KE2\)

In elastic collisions, both momentum and kinetic energies are conserved. We can use the momentum conservation principle to find the relative velocities of the pendulum and the block after collision. Let the velocity of mass m1 (pendulum) be \(v1_f\) and the velocity of mass m2 (block) be \(v2_f\) after collision. The initial velocities are 0. The equation for conservation of momentum is: \(m1 * v1_i + m2 * v2_i = m1 * v1_f + m2 * v2_f\) As both masses are equal and initial velocities are zero, we get: \(m * 0 + m * 0 = m * v1_f + m * v2_f\) Solving for \(v2_f\): \(v2_f = -v1_f\)

Now we have the relationship between the final velocities of the two masses, we can use it to find the kinetic energy of the block (m2). From step 4, we have: \(v2_f = -v1_f\) The kinetic energy of a mass is given by: \(KE = \frac{1}{2}mv^2\) Substitute the expression for \(v2_f\) from step 4: \(KE2 = \frac{1}{2}m(-v1_f)^2\) \(KE2 = \frac{1}{2}mv1_f^2\) Now we can substitute this expression for KE2 in the energy conservation equation from step 3: \(mg = KE1 + \frac{1}{2}mv1_f^2\) We also know that: \(KE1 = \frac{1}{2}mv1_f^2\) So the equation becomes: \(mg = 2 * \frac{1}{2}mv1_f^2\) Divide both sides by m to obtain the Kinetic Energy of the block (KE2): \(KE2 = 2g * 1\) \(KE2 = 2mg\) Hence, the kinetic energy of the block after the collision is \(2mg\). So, the correct answer is: (A) \(2mg1\)