To solve the problem of a cubical block of mass \( m \) sliding down a movable wedge of mass \( M \), we need to apply the principles of conservation of momentum and energy. Let’s break this down step by step to find the velocity of the wedge when the block reaches the bottom.
Understanding the System
We have a two-body system: the cubical block and the wedge. The block is released from a height \( h \) and slides down the wedge, which is also free to move on a frictionless surface. Since both surfaces are frictionless, we can assume that there are no external horizontal forces acting on the system.
Applying Conservation of Momentum
Initially, both the block and the wedge are at rest. When the block starts sliding down, the wedge will move in the opposite direction due to the conservation of momentum. The total momentum of the system before the block is released is zero:
When the block reaches the bottom of the wedge, let \( v \) be the velocity of the block relative to the ground, and \( V \) be the velocity of the wedge. The momentum of the system at this point can be expressed as:
- Final momentum = \( mv - MV \)
Setting the initial momentum equal to the final momentum gives us:
0 = \( mv - MV \)
From this, we can derive the relationship:
mv = MV
or
v = \frac{M}{m} V
Using Conservation of Energy
Next, we apply the principle of conservation of energy. The potential energy of the block at height \( h \) is converted into kinetic energy as it slides down the wedge. The initial potential energy \( PE \) of the block is given by:
As the block reaches the bottom, it has kinetic energy \( KE \) given by:
- Final \( KE = \frac{1}{2} mv^2 + \frac{1}{2} MV^2 \)
Setting the initial potential energy equal to the total kinetic energy gives us:
mgh = \( \frac{1}{2} mv^2 + \frac{1}{2} MV^2 \)
Substituting for Velocities
Now, we can substitute \( v \) from our momentum equation into the energy equation:
mgh = \( \frac{1}{2} m \left(\frac{M}{m} V\right)^2 + \frac{1}{2} MV^2 \)
This simplifies to:
mgh = \( \frac{1}{2} \frac{M^2}{m} V^2 + \frac{1}{2} MV^2 \)
Combining the terms gives:
mgh = \( \frac{1}{2} V^2 \left(\frac{M^2}{m} + M\right) \)
Solving for Velocity
Now, we can solve for \( V \):
V^2 = \frac{2mgh}{\frac{M^2}{m} + M}
V = \sqrt{\frac{2mgh}{\frac{M^2}{m} + M}}
Finally, substituting this back into our equation for \( v \):
v = \frac{M}{m} V = \frac{M}{m} \sqrt{\frac{2mgh}{\frac{M^2}{m} + M}}
Final Expression
Thus, the velocity of the cubical block when it reaches the bottom of the wedge is:
v = \frac{M}{m} \sqrt{\frac{2mgh}{\frac{M^2}{m} + M}}
This equation gives you the velocity of the block in terms of the masses \( m \) and \( M \), the height \( h \), and the gravitational acceleration \( g \). It illustrates how the motion of one body affects the other in a system where both are free to move.
