91Ó°ÊÓ

Initially, both body P and block Q are at rest. So, the initial velocities of both P and Q are zero.

As there is no external force acting in the horizontal direction, we can apply the conservation of linear momentum principle. In other words, the total linear momentum of the system before and after body P slides down the slope should be the same. Let's denote the final velocities of body P and block Q as \(v_P\) and \(v_Q\) respectively. Initial momentum of the system = Final momentum of the system \(0 = m v_P - M v_Q\)

As body P slides down the inclined slope, we need to find the horizontal component of its final velocity. This can be done using basic trigonometry. If we assume no energy is lost, the gravitational force does the work of pulling body P down the slope. The horizontal component of the final velocity of body P can be given by: \(v_P = v_{Px} \cos \theta \)

Now we can substitute the horizontal component of body P's velocity into the momentum conservation equation from Step 2: \(0 = m(v_{Px} \cos \theta) - M v_Q\) Now let's find a relationship between the horizontal component of body P's velocity and the distance traveled by block Q. As body P slides down a length L, we can express the horizontal distance traveled by body P: \(x = L \cos \theta\) Now we have the relationship: \(v_{Px} = \frac{x}{t}\), where t is the time taken. Substitute this expression into the momentum conservation equation: \(0 = m(\frac{x}{t} \cos \theta) - M v_Q\) Now we can express the displacement of block Q in terms of L and the mass ratio: \(x = M \frac{mv_Q}{M+m} t = M \frac{mv_Q}{M+m}\frac{x}{L \cos \theta}\) Finally, we have: \(x = \frac{m L \cos \theta}{m+M}\) The correct answer is (D), \(\frac{m L \cos \theta}{m+M}\).