© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Understand the Physical Setup
A wedge of mass $M=4m$ is placed on a frictionless horizontal plane. A particle of mass $m$ is moving toward it with speed $v$. All surfaces are frictionless, so no energy is lost due to friction. We want to determine the maximum height $h$ the particle reaches on the wedge.
The initial configuration can be visualized as shown:
Step 2: Conservation of Linear Momentum (Horizontal Direction)
Just after the particle contacts the wedge, suppose both the wedge and the particle move together momentarily with some common horizontal velocity $v'$. Since the plane is frictionless, the horizontal momentum of the system is conserved.
Before collision:
Particle has momentum $mv$ (to the right).
Wedge is at rest.
After collision (immediately):
Combined mass is $m + 4m = 5m$ moving with speed $v'$.
Therefore,
$$ mv = (m + 4m)\,v' = 5m\,v' $$
which simplifies to
$$ v' = \frac{v}{5}. $$
Step 3: Conservation of Mechanical Energy
Use the conservation of energy between the initial state (particle moving with speed $v$, wedge at rest) and the final state (when the particle has climbed to its maximum height $h$).
Initially, total kinetic energy is
$$ \frac{1}{2}m v^2. $$
Finally, at maximum height $h$, the wedge and the particle have some kinetic energy (because the wedge can move freely) plus the potential energy of the particle at height $h$.
So, we write
$$
\frac{1}{2} m v^2
= \frac{1}{2}(m + 4m)\,{v'}^2 + m g h \,.
$$
Substituting $v' = \frac{v}{5}$, we get
$$
\frac{1}{2} m v^2
= \frac{1}{2}\,(5m)\,\left(\frac{v}{5}\right)^2 + m g h \,.
$$
Simplifying,
$$
\frac{1}{2} m v^2
= \frac{1}{2} \times 5m \times \frac{v^2}{25} + m g h \,,
$$
$$
\frac{1}{2} m v^2
= \frac{1}{2} m \frac{v^2}{5} + m g h \,.
$$
Thus,
$$
\frac{1}{2} m v^2 - \frac{1}{2} m \frac{v^2}{5}
= m g h \,.
$$
$$
\frac{1}{2} m v^2 \left(1 - \frac{1}{5}\right)
= m g h \,.
$$
$$
\frac{1}{2} m v^2 \times \frac{4}{5}
= m g h \,.
$$
$$
\frac{2}{5} m v^2
= m g h \,.
$$
Canceling $m$ on both sides:
$$
h = \frac{2 v^2}{5 g}.
$$
We thus see that the particle rises to a maximum height of
$$ h = \frac{2 v^2}{5 g}. $$
Therefore, the correct answer is:
$$ \frac{2 v^2}{5 g}. $$
