A wedge of mass m and triangular cross-section (AB = BC = CA = 2R) is moving with a constant velocit

Q: Wait, why does the wedge rebound with the exact same speed $v$?

Because the collision is perfectly elastic (kinetic energy is conserved) and the sphere is fixed to the table (its mass is effectively infinite). In such a head-on elastic collision with an immovable wall, an object simply reverses its velocity.

Q: How did we figure out that the force $\mathbf{F}$ is at a $30^\circ$ angle?

The wedge is an equilateral triangle, so its face makes a $60^\circ$ angle with the horizontal table. The force $\mathbf{F}$ exerted by the smooth sphere acts along the common normal, which is perpendicular to this face. A line perpendicular to a $60^\circ$ incline makes a $30^\circ$ angle with the horizontal.

Q: Why do we include $mg$ when calculating the normal force during a collision?

Usually, non-impulsive forces like gravity are ignored during a collision because they are much smaller than the impulsive forces. However, the question specifically asks for the *total* normal force exerted by the table during the interval $\Delta t$. This total force must balance both the impulsive downward push from the sphere and the wedge's own weight.

Q: Why is the torque due to gravity zero?

Torque is defined as force times the perpendicular distance from the pivot point to the line of action of the force. Since we are calculating torque about the Center of Mass (CM), and gravity acts exactly *at* the CM, the perpendicular distance is zero. Hence, no torque!

Q: Why must the torque from $\mathbf{N}$ equal the torque from $\mathbf{F}$?

The problem states the wedge returns "without any rotation". This means its angular acceleration is zero, which by Newton's Second Law for rotation ($\tau_{net} = I\alpha$), means the net torque about the CM must be zero. Since gravity provides no torque, the clockwise torque from $\mathbf{N}$ must perfectly cancel the counter-clockwise torque from $\mathbf{F}$.

Question

A wedge of mass $m$ and triangular cross-section ($AB = BC = CA = 2R$) is moving with a constant velocity ($-v\hat{i}$) towards a sphere of radius $R$ fixed on a smooth horizontal table as shown in the figure.
The wedge makes an elastic collision with the fixed sphere and returns along the same path without any rotation. Neglect all friction and suppose that the wedge remains in contact with the sphere for a very short time $\Delta t$ during which the sphere exerts a constant force $\mathbf{F}$ on the wedge.
(a) Find the force $\mathbf{F}$ and also the normal force $\mathbf{N}$ exerted by the table on the wedge during the time $\Delta t$.
(b) Let $h$ denote the perpendicular distance between the centre of mass of the wedge and the line of action of $F$. Find the magnitude of the torque due to the normal force $\mathbf{N}$ about the centre of the wedge during the interval $\Delta t$.

Sigma Percentile · Accepted Answer

This problem is a beautiful symphony of linear momentum, impulsive forces, and rotational equilibrium. It tests your ability to break down a complex, instantaneous event into manageable physical principles. Let's dive into the mechanics of this elastic collision.

### The Elastic Rebound

Imagine the wedge sliding smoothly across the table with an initial velocity of $-v\hat{i}$. It crashes into a fixed, immovable sphere. The problem explicitly states that this is an **elastic collision**.

Because the sphere is fixed to the table (effectively having infinite mass), it absorbs no kinetic energy. The wedge simply bounces off the sphere, reversing its direction while maintaining its original speed. Therefore, the final velocity of the wedge is $+v\hat{i}$.

This gives us a massive change in horizontal momentum:

$$ \Delta p_x = p_{xf} - p_{xi} = mv - (-mv) = 2mv $$

### Unpacking the Collision Force

This change in momentum doesn't happen by magic; it is the direct result of the impulsive force $\mathbf{F}$ exerted by the sphere on the wedge during the tiny time interval $\Delta t$.

Since the sphere is smooth, this force acts along the common normal at the point of contact. The wedge is an equilateral triangle, meaning its face is inclined at $60^\circ$ to the horizontal. A line perpendicular to a $60^\circ$ incline makes a $30^\circ$ angle with the horizontal. Thus, the force $\mathbf{F}$ points downwards and to the right at an angle of $30^\circ$.

According to the Impulse-Momentum Theorem, the horizontal impulse equals the change in horizontal momentum:

$$ (F \cos 30^\circ) \Delta t = 2mv $$

Substituting $\cos 30^\circ = \frac{\sqrt{3}}{2}$, we can solve for the magnitude of the force:

$$ F = \frac{4mv}{\sqrt{3}\Delta t} $$

To express this as a vector, we break it into its $x$ and $z$ components:

$$ \mathbf{F} = (F \cos 30^\circ)\hat{i} - (F \sin 30^\circ)\hat{k} $$

$$ \mathbf{F} = \frac{2mv}{\Delta t}\hat{i} - \frac{2mv}{\sqrt{3}\Delta t}\hat{k} $$

### Vertical Equilibrium and the Normal Force

During the collision, the wedge doesn't fly up into the air or sink into the table. It remains in strict vertical equilibrium. This means the upward normal force $\mathbf{N}$ from the table must perfectly balance all downward forces.

What are the downward forces? There is the wedge's own weight, $mg$, and the downward component of the collision force, $F \sin 30^\circ$.

$$ N = mg + F \sin 30^\circ $$

Substituting our known value for $F$ and $\sin 30^\circ = \frac{1}{2}$:

$$ N = mg + \frac{2mv}{\sqrt{3}\Delta t} $$

In vector form, since it points straight up:

$$ \mathbf{N} = \left(mg + \frac{2mv}{\sqrt{3}\Delta t}ight)\hat{k} $$

### The No-Rotation Constraint

The final part of the problem asks for the torque due to the normal force. The golden key here is the phrase: **"returns along the same path without any rotation."**

If the wedge doesn't rotate, its angular acceleration is zero. By Newton's Second Law for rotation, the net torque about its Center of Mass (CM) must be exactly zero.

$$ 	au_{net} = 	au_N + 	au_F + 	au_{mg} = 0 $$

Gravity acts exactly at the CM, so its lever arm is zero, meaning $	au_{mg} = 0$. This leaves us with a simple balance: the magnitude of the torque from the normal force must equal the magnitude of the torque from the collision force.

$$ |	au_N| = |	au_F| $$

We are given that the perpendicular distance from the CM to the line of action of $\mathbf{F}$ is $h$. Therefore, the torque from $\mathbf{F}$ is simply $F \cdot h$.

$$ |	au_N| = \left(\frac{4mv}{\sqrt{3}\Delta t}ight)h $$

And just like that, by trusting the physical constraints of the system, the answer reveals itself elegantly!

Visualized Solution

Visualizing the Collision

Change in Horizontal Momentum

Geometry of the Collision Force

Impulse-Momentum Theorem

Vector Form of Force $F$

Vertical Equilibrium

Calculating Normal Force $N$

The No-Rotation Constraint

Torque due to Normal Force

The Sigma Insight: Head-on Collision

The Elastic Rebound

Unpacking the Collision Force

Vertical Equilibrium and the Normal Force

The No-Rotation Constraint

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Visualized Solution

Visualizing the Collision

Change in Horizontal Momentum

Geometry of the Collision Force

Impulse-Momentum Theorem

Vector Form of Force .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F

Vertical Equilibrium

Calculating Normal Force .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N

The No-Rotation Constraint

Torque due to Normal Force

The Sigma Insight: Head-on Collision

The Elastic Rebound

Unpacking the Collision Force

Vertical Equilibrium and the Normal Force

The No-Rotation Constraint

Similar Questions

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