91Ó°ÊÓ

Whenever a system is isolated, like the one in our exercise where no external horizontal forces act on plank 'A' and block 'B', the principle of conservation of momentum comes into play. Initially, only block 'B' is moving with a velocity 'u', so it has all the system's momentum. The initial momentum is given by \[ \frac{m}{2} u, \] where \( m \) is the mass of the plank. As the system reaches a common velocity \( v_f = \frac{u}{3} \), both the block and plank move together, leading to a final momentum of \[ \left(m + \frac{m}{2}\right) v_f. \] By equating initial and final momentum, we confirm that momentum is conserved, leading to the equation:\[ \frac{m}{2} u = \frac{3m}{2} v_f. \] By solving this, \( v_f \) is verified to be \( \frac{u}{3} \). This illustrates the system's constant momentum and demonstrates the dynamic relationship between both masses as they adjust to move together.

Kinetic energy is the energy of motion. Initially, block 'B' has kinetic energy due to its motion given by:\[ \frac{1}{2} \cdot \frac{m}{2} \cdot u^2 = \frac{mu^2}{4}. \]This energy enables block 'B' to move across surface 'A'.As friction acts and both the block and plank A reach a common speed, their combined kinetic energy is calculated as:\[ \text{K.E. final} = \frac{1}{2} \cdot \frac{3m}{2} \cdot \left(\frac{u}{3}\right)^2 = \frac{mu^2}{12}. \]The reduction from initial to final kinetic energy (\( \frac{mu^2}{4} \to \frac{mu^2}{12} \))is due to the work done by friction. We calculated the work done as:\[ \frac{mu^2}{6}, \]which matches two-thirds of the initial kinetic energy. This helps us understand how energy transitions from full motion to overcoming opposing forces like friction.

Friction is the resisting force that acts between surfaces in contact. In this exercise, the coefficient of friction \( \mu \) triggers changes in the dynamics of both block 'B' and plank 'A'. The force of friction is given by:\[ F = \mu \frac{m}{2} g, \]where \( g \) is the acceleration due to gravity.This frictional force impacts motion by decelerating block 'B' and accelerating plank 'A'. Consequently, the acceleration for plank 'A' is:\[ a_A = \frac{\mu g}{2}. \]Moreover, friction causes a loss in kinetic energy, as described in the previous section, leading to the energy being converted into other forms, like heat. Through these changes, we see how friction can drastically influence overall kinetics.

Relative motion involves understanding how the motions of two bodies compare. Before plank 'A' and block 'B' achieve a common velocity, their relative accelerations determine how their speeds change compared to each other. The force of friction acts differently on both, causing separate accelerations.The block 'B' experiences deceleration \( a_B = -\mu g \), while plank 'A' is accelerated with \( a_A = \frac{\mu g}{2} \). The relative acceleration is then the magnitude of the difference between their accelerations:\[ |a_{BA}| = |a_B - a_A| = |(-\mu g) - (\frac{\mu g}{2})| = \frac{3}{2} \mu g, \]confirming the construction of relative motion's effect in the system. This shows how internal forces (i.e., friction) can cause shifts and adjustments in the motion of interconnected bodies.