To tackle this problem, we need to analyze the situation step by step, considering the principles of conservation of momentum and the conditions for pure rolling. Let's break it down into manageable parts.
Understanding the System
We have a block of mass M moving with an initial velocity \( v_0 \) that collides elastically with a plank of the same mass M, which is initially at rest. On this plank, there is a solid cylinder of mass M also at rest. The coefficient of friction between the plank and the cylinder is \( x \). The goal is to find the velocities of the center of mass of the cylinder when it starts pure rolling on the plank and the velocity of the center of mass of the plank when the cylinder starts pure rolling.
Step 1: Analyzing the Collision
Since the collision between the block and the plank is perfectly elastic, we can use the conservation of momentum and kinetic energy. Before the collision, the total momentum is:
- Initial momentum = \( Mv_0 + 0 = Mv_0 \)
After the collision, let the velocity of the block be \( v_1 \) and the velocity of the plank be \( v_p \). The conservation of momentum gives us:
For a perfectly elastic collision, the kinetic energy is also conserved:
- \( \frac{1}{2}Mv_0^2 = \frac{1}{2}Mv_1^2 + \frac{1}{2}Mv_p^2 \)
Step 2: Solving for Velocities After Collision
From the momentum equation, we can express \( v_p \) in terms of \( v_1 \):
Substituting this into the kinetic energy equation allows us to solve for \( v_1 \) and \( v_p \). After some algebra, we find:
- \( v_1 = \frac{v_0}{2} \)
- \( v_p = \frac{v_0}{2} \)
Step 3: Cylinder Starts Pure Rolling
Now, we need to consider the cylinder on the plank. The cylinder will start pure rolling when the frictional force between the plank and the cylinder is sufficient to prevent slipping. The condition for pure rolling is that the linear acceleration of the center of mass of the cylinder equals the angular acceleration times the radius of the cylinder.
Frictional Force and Acceleration
The frictional force \( f \) acting on the cylinder is given by:
Where \( \mu \) is the coefficient of friction \( x \) and \( g \) is the acceleration due to gravity. The linear acceleration \( a \) of the cylinder is:
- \( a = \frac{f}{M} = \frac{xMg}{M} = xg \)
The angular acceleration \( \alpha \) of the cylinder is related to the linear acceleration by:
- \( \alpha = \frac{a}{R} = \frac{xg}{R} \)
Velocity of the Cylinder's Center of Mass
When the cylinder starts rolling without slipping, its linear velocity \( v_c \) can be expressed in terms of the distance traveled and time. Assuming it starts from rest, we can relate the velocities:
- \( v_c = R\alpha t = R\left(\frac{xg}{R}\right)t = xgt \)
At the moment it starts pure rolling, the velocity of the cylinder's center of mass is equal to the velocity of the plank:
- \( v_c = v_p = \frac{v_0}{2} \)
Thus, we can find the time \( t \) when this occurs, leading us to the velocity of the center of mass of the cylinder when it starts pure rolling:
- \( v_c = \frac{v_0}{2} \)
Step 4: Velocity of the Plank
Since the plank is also moving with velocity \( v_p \) when the cylinder starts rolling, we can conclude:
- \( v_p = \frac{v_0}{2} \)
Final Results
In summary, the velocities are:
- Velocity of the center of mass of the cylinder when it starts pure rolling: \( \frac{v_0}{2} \)
- Velocity of the center of mass of the plank when the cylinder starts pure rolling: \( \frac{v_0}{2} \)
This analysis shows how the principles of momentum and energy conservation, along with the conditions for rolling motion, come together to provide a clear understanding of the dynamics involved in this scenario.
