A wedge of mass M=4m lies on a frictionless plane. A particle of mass m approaches the wedge with sp

Q: Why do we assume the particle and wedge have the same velocity at maximum height?

At maximum height, the particle stops moving *relative* to the wedge. If it had a different horizontal velocity, it would either still be moving up the incline or sliding back down. Thus, their velocities must be identical at that instant.

Q: Why isn't momentum conserved in the vertical direction?

Momentum is only conserved when the net external force is zero. In the vertical direction, gravity acts downwards on both masses, and the normal force from the ground acts upwards. These external forces change the vertical momentum of the system.

Q: Could we solve this using the work-energy theorem instead of conservation of energy?

Yes! The work-energy theorem states that the work done by all forces equals the change in kinetic energy. Since the only force doing work is gravity (a conservative force), the work done by gravity is exactly the negative change in potential energy, which mathematically leads to the exact same conservation of mechanical energy equation.

Q: What would happen if there was friction between the particle and the wedge?

If friction were present, it would do negative work on the system, converting some of the mechanical energy into heat. In that case, mechanical energy would not be conserved, and the maximum height reached would be lower than $\frac{2v^2}{5g}$.

Q: Why did we use $5m$ in the final kinetic energy term?

At the maximum height, both the particle (mass $m$) and the wedge (mass $4m$) are moving together at the same speed $v'$. Therefore, the total moving mass is $m + 4m = 5m$, and their combined kinetic energy is $\frac{1}{2}(5m)v'^2$.

Question

A wedge of mass $M=4m$ lies on a frictionless plane. A particle of mass $m$ approaches the wedge with speed $v$. There is no friction between the particle and the plane or between the particle and the wedge. The maximum height climbed by the particle on the wedge is given by

Options:

(A) $\frac{2v^2}{7g}$

(B) $\frac{v^2}{g}$

(C) $\frac{2v^2}{5g}$

(D) $\frac{v^2}{2g}$

Sigma Percentile · Accepted Answer

### The Setup and the Catch

Imagine a particle of mass $m$ sliding on a perfectly smooth horizontal surface with a velocity $v$. Ahead of it lies a wedge of mass $4m$.

If the wedge were bolted to the ground, this would be a simple problem of converting kinetic energy into gravitational potential energy. But here is the catch: the wedge is free to move!

Because there is absolutely no friction, the moment the particle starts climbing the wedge, it pushes the wedge forward. This means some of the particle's initial kinetic energy will be "stolen" to give the wedge kinetic energy.

### The Concept of Maximum Height

What exactly happens at the "maximum height"?

As the particle climbs, it slows down relative to the wedge, while the wedge speeds up. At the exact moment the particle reaches its highest point on the wedge, its vertical velocity becomes zero.

More importantly, its horizontal velocity becomes exactly equal to the velocity of the wedge. If it were moving faster than the wedge, it would still be climbing. If it were moving slower, it would be sliding back down.

Therefore, at maximum height, both the particle and the wedge move together as a single unit with a common horizontal velocity, let's call it $v'$.

### Conservation of Linear Momentum

Since there are no external forces acting on the system in the horizontal direction (gravity acts downwards, and normal forces are vertical), the horizontal momentum of the system is strictly conserved.

**Initial Momentum = Final Momentum**

Before the collision, only the particle is moving:
$$p_i = mv$$

At maximum height, both masses move together with velocity $v'$:
$$p_f = (m + 4m)v' = 5mv'$$

Equating the two, we can easily find the common velocity:
$$mv = 5mv' \implies v' = \frac{v}{5}$$

### Conservation of Mechanical Energy

Now, let's track the energy. Since all surfaces are frictionless, no energy is lost to heat or sound. The total mechanical energy of the system remains constant.

**Initial Energy = Final Energy**

Initially, the system only has the kinetic energy of the particle:
$$E_i = \frac{1}{2}mv^2$$

At maximum height $h$, the system has the kinetic energy of both masses moving at $v'$, plus the gravitational potential energy of the particle:
$$E_f = \frac{1}{2}(m + 4m)v'^2 + mgh$$

Equating the initial and final energies:
$$\frac{1}{2}mv^2 = \frac{1}{2}(5m)v'^2 + mgh$$

### The Final Calculation

Let's substitute the value of $v' = \frac{v}{5}$ into our energy equation:
$$\frac{1}{2}mv^2 = \frac{1}{2}(5m)\left(\frac{v}{5}\right)^2 + mgh$$

Squaring the velocity term:
$$\frac{1}{2}mv^2 = \frac{1}{2}(5m)\left(\frac{v^2}{25}\right) + mgh$$

Simplifying the kinetic energy term on the right:
$$\frac{1}{2}mv^2 = \frac{mv^2}{10} + mgh$$

Now, we can cancel the mass $m$ from every term and rearrange to solve for $h$:
$$gh = \frac{v^2}{2} - \frac{v^2}{10}$$

Finding a common denominator:
$$gh = \frac{5v^2 - v^2}{10} = \frac{4v^2}{10} = \frac{2v^2}{5}$$

Finally, isolating $h$:
**$$h = \frac{2v^2}{5g}$$**

This is our final answer! Notice that if the wedge were fixed, the height would have been $\frac{v^2}{2g}$. Because the wedge was free to move, it carried away some kinetic energy, resulting in a lower maximum height.

Select Answer:

Visualized Solution

$m, 4 m, v$

$v_{r e l} = 0 ⟹ v_{co mm o n} = v^{'}$

$F_{e x t, x} = 0 ⟹ Δ p_{x} = 0$

$p_{i} = p_{f} ⟹ v^{'} = \frac{v}{5}$

$W_{n c} = 0 ⟹ Δ E = 0$

$K_{i} = K_{f} + U_{f}$

$\frac{1}{2} m v^{2} = \frac{1}{2} (5 m) v^{'2} + m g h$

$\frac{1}{2} m v^{2} = \frac{1}{2} (5 m) (\frac{v}{5})^{2} + m g h$

$\frac{1}{2} m v^{2} = \frac{m v ^{2}}{10} + m g h$

$h = \frac{2 v ^{2}}{5 g}$

$h_{f i x e d} = \frac{v ^{2}}{2 g} > \frac{2 v ^{2}}{5 g}$

The Sigma Insight: Conservation of Linear Momentum

The Setup and the Catch

The Concept of Maximum Height

Conservation of Linear Momentum

Conservation of Mechanical Energy

The Final Calculation

Similar Questions

Two particles of equal mass $m$ have respective initial velocities $u \hat{i}$ and $u (\frac{i ^ + j ^}{2})$ . They collide completely inelastically. The energy lost in the process is

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Visualized Solution

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