A block of mass 𝑚 m slides down a frictionless incline of angle 𝜃 θ and then collides elastically with a stationary block of mass 𝑀 M at the bottom. Calculate the final velocity of both blocks after the collision in terms of 𝑚 m, 𝑀 M, and
To solve this problem, we will break it down into two parts: first, we will find the velocity of the block mass \( m \) just before the collision, and then we will use the principles of momentum conservation and kinetic energy conservation to find the final velocities after the elastic collision.
### Step 1: Determine velocity \( v \) of block \( m \) just before collision
A block of mass \( m \) slides down a frictionless incline of angle \( \theta \). The potential energy at the top of the incline is converted into kinetic energy just before it collides with the stationary block of mass \( M \).
Let \( h \) be the height of the incline. The height can be expressed in terms of the length of the incline \( L \) as:
\[
h = L \sin(\theta)
\]
The gravitational potential energy at the height \( h \) is given by:
\[
PE = mgh
\]
where \( g \) is the acceleration due to gravity. The kinetic energy just before the collision is:
\[
KE = \frac{1}{2} mv^2
\]
By the conservation of energy:
\[
mgh = \frac{1}{2} mv^2
\]
Substituting for \( h \):
\[
mg(L \sin(\theta)) = \frac{1}{2} mv^2
\]
Cancelling \( m \) (assuming \( m \neq 0 \)):
\[
gL \sin(\theta) = \frac{1}{2} v^2
\]
Solving for \( v \):
\[
v^2 = 2gL \sin(\theta)
\]
\[
v = \sqrt{2gL \sin(\theta)}
\]
### Step 2: Collision analysis
Now we will analyze the elastic collision between the block of mass \( m \) moving with velocity \( v \) and the stationary block of mass \( M \).
**Conservation of Momentum:**
The total momentum before the collision must equal the total momentum after the collision:
\[
mv = mv_f + MV_F
\]
where
- \( v_f \) is the final velocity of block \( m \),
- \( V_F \) is the final velocity of block \( M \).
**Conservation of Kinetic Energy:**
The total kinetic energy before the collision must equal the total kinetic energy after the collision:
\[
\frac{1}{2} mv^2 = \frac{1}{2} mv_f^2 + \frac{1}{2} MV_F^2
\]
### Solving the equations
By using the equations above, and knowing that in an elastic collision the velocities can be calculated using specific formulas can simplify this process. For a perfectly elastic collision, for masses \( m \) and \( M \):
\[
v_f = \frac{(m - M)}{(m + M)} v \quad \text{(velocity of m after collision)}
\]
\[
V_F = \frac{(2m)}{(m + M)} v \quad \text{(velocity of M after collision)}
\]
### Final expressions
Substituting \( v = \sqrt{2gL \sin(\theta)} \) gives:
1. The final velocity of block \( m \):
\[
v_f = \frac{(m - M)}{(m + M)} \sqrt{2gL \sin(\theta)}
\]
2. The final velocity of block \( M \):
\[
V_F = \frac{2m}{(m + M)} \sqrt{2gL \sin(\theta)}
\]
This gives us the final velocities of both blocks \( m \) and \( M \) after the elastic collision in terms of \( m \), \( M \), and \( \theta \).