To tackle this problem, we need to break it down into two parts: first, we will find the distance along the bottom of the box between the point of projection P and the point Q where the particle lands. Then, we will determine the speed of the box with respect to the ground when the particle is projected. Let's dive into the details.
Part (a): Finding the Distance Between Points P and Q
We start by analyzing the motion of the particle relative to the box. The particle is projected with a speed \( u \) at an angle \( \alpha \) relative to the bottom of the box. Since the box is sliding down the incline without friction, we can use the principles of projectile motion to find the distance traveled by the particle inside the box.
Step 1: Resolve the Velocity Components
The initial velocity of the particle can be broken down into two components:
- The horizontal component (along the bottom of the box): \( u_x = u \cos(\alpha) \)
- The vertical component (perpendicular to the bottom of the box): \( u_y = u \sin(\alpha) \)
Step 2: Determine the Time of Flight
Next, we need to find the time the particle remains in the air. The vertical motion is influenced by gravity. The vertical displacement \( y \) of the particle can be described by the equation:
\( y = u_y t - \frac{1}{2} g t^2 \)
Since the particle does not hit any other surface of the box, it will land back on the bottom of the box, meaning the vertical displacement \( y \) is equal to the height of the box at the moment of projection. If we denote the height of the box as \( h \), we can set \( y = h \) and solve for \( t \):
\( h = u \sin(\alpha) t - \frac{1}{2} g t^2 \)
Step 3: Horizontal Displacement
During the time \( t \), the horizontal displacement \( d \) of the particle relative to the box is given by:
\( d = u_x t = u \cos(\alpha) t \)
Step 4: Relating the Time and Distance
To find the distance \( d \) along the bottom of the box, we need to express \( t \) in terms of \( h \) and the initial velocities. We can rearrange the vertical motion equation to solve for \( t \) and substitute it back into the horizontal displacement equation. This will yield:
\( d = \frac{u \cos(\alpha)}{g} \left( u \sin(\alpha) + \sqrt{(u \sin(\alpha))^2 + 2gh} \right) \)
Part (b): Speed of the Box with Respect to the Ground
Now, we need to find the speed of the box with respect to the ground at the moment the particle is projected, given that the horizontal displacement of the particle with respect to the ground is zero. This means that the horizontal motion of the box and the particle must cancel each other out.
Step 1: Analyze the Horizontal Motion
Let \( v \) be the speed of the box with respect to the ground. The horizontal motion of the particle with respect to the ground can be expressed as:
\( v_{particle} = v + u \cos(\alpha) \)
Step 2: Set Horizontal Displacement to Zero
Since the horizontal displacement of the particle with respect to the ground is zero, we have:
\( v + u \cos(\alpha) = 0 \)
Step 3: Solve for the Speed of the Box
From this equation, we can solve for \( v \):
\( v = -u \cos(\alpha) \)
This negative sign indicates that the box is moving in the opposite direction to the horizontal component of the particle's velocity. Thus, the speed of the box with respect to the ground at the moment of projection is:
\( v = u \cos(\alpha) \) (in the direction opposite to the particle's horizontal motion)
Summary
In summary, we found that the distance along the bottom of the box between points P and Q is given by the derived expression involving the initial speed \( u \), angle \( \alpha \), and height \( h \). Additionally, the speed of the box with respect to the ground at the moment of projection is \( u \cos(\alpha) \), moving in the opposite direction to the particle's horizontal component. This analysis highlights the interplay between the motion of the box and the particle within it, illustrating fundamental concepts of projectile motion and relative velocity.
