A particle of mass m is moving with speed 2v and collides with a mass 2m moving with speed v in the

Q: Why don't we conserve kinetic energy here?

The problem states that the second mass splits into two particles, and the first mass stops. This is a completely inelastic collision (or an explosion-like event), where kinetic energy is not conserved. Only linear momentum is conserved when no external forces act on the system.

Q: Why did we only consider the x-direction for momentum conservation?

We could consider the y-direction too! However, the initial momentum in the y-direction is zero. The final momentum in the y-direction is $mv'\sin 45^\circ - mv'\sin 45^\circ = 0$, which is automatically satisfied and doesn't help us find $v'$. The x-direction equation contains all the information we need.

Q: What happens to the first mass $m$?

The problem explicitly states that the first mass is 'stopped completely' after the collision. This means its final velocity is zero, so it contributes nothing to the final momentum.

Q: Why is the mass of the split particles $m$ each?

The original second particle had a mass of $2m$. The problem states it 'splits into two particles each of mass $m$'. This perfectly conserves the mass of that specific object ($m + m = 2m$).

Q: How did $4 / \sqrt{2}$ become $2\sqrt{2}$?

It's a simple rationalization. $4$ can be written as $2 \times 2$, and $2$ is $\sqrt{2} \times \sqrt{2}$. So, $4 / \sqrt{2} = (2 \times \sqrt{2} \times \sqrt{2}) / \sqrt{2} = 2\sqrt{2}$.

Question

A particle of mass $m$ is moving with speed $2v$ and collides with a mass $2m$ moving with speed $v$ in the same direction. After collision, the first mass is stopped completely while the second one splits into two particles each of mass $m$, which move at angle $45^{\circ}$ with respect to the original direction. The speed of each of the moving particle will be

Options:

(A) $\sqrt{2} v$

(B) $\frac{v}{\sqrt{2}}$

(C) $\frac{v}{2\sqrt{2}}$

(D) $2\sqrt{2} v$

Sigma Percentile · Accepted Answer

### The Collision Setup

Let's visualize the situation before the collision. We have a particle of mass $m$ moving with a speed of $2v$. Ahead of it, there is another particle of mass $2m$ moving in the exact same direction with a speed of $v$. Because the trailing particle is moving faster, a collision is inevitable.

### The Aftermath

Now, what happens after the collision? The problem gives us a very specific set of conditions. The first mass $m$ comes to a complete halt; its final velocity is zero. The second mass $2m$ undergoes a dramatic change—it splits into two equal parts, each of mass $m$.

These two new particles don't just keep moving straight. They fly off at $45^\circ$ angles relative to the original direction of motion. Let's call their new speed $v'$. Because the setup is perfectly symmetric, both particles will have the same speed $v'$.

### The Power of Momentum Conservation

Since there are no external forces acting on this system of particles, the total linear momentum must be conserved. This is a fundamental law of physics. We can apply this principle along the original direction of motion, which we'll designate as the x-axis.

Let's set up our master equation. The total initial momentum in the x-direction must equal the total final momentum in the x-direction:

$$p_{ix} = p_{fx}$$

### Crunching the Numbers

Calculating the initial momentum is straightforward. We simply sum the momentum of the two particles before the crash:

$$p_{ix} = m(2v) + 2m(v)$$

$$p_{ix} = 2mv + 2mv = 4mv$$

For the final momentum, we have three pieces to consider. The first mass has zero velocity, so its momentum is zero. The two new masses each have an x-component of velocity equal to $v' \cos 45^\circ$. Let's add them up:

$$p_{fx} = m(0) + m(v' \cos 45^\circ) + m(v' \cos 45^\circ)$$

$$p_{fx} = 2mv' \cos 45^\circ$$

We know from trigonometry that $\cos 45^\circ = \frac{1}{\sqrt{2}}$. Substituting this into our equation, the final momentum simplifies beautifully:

$$p_{fx} = 2mv' \left(\frac{1}{\sqrt{2}}ight) = \sqrt{2}mv'$$

### The Final Result

Now, we bring it all together by equating the initial and final momentums:

$$4mv = \sqrt{2}mv'$$

We can cancel the mass $m$ from both sides, as it's a common factor. Now, we just solve for our unknown speed, $v'$:

$$v' = \frac{4v}{\sqrt{2}}$$

To simplify this, remember that $4$ can be written as $2 	imes 2$, and $2$ is $\sqrt{2} 	imes \sqrt{2}$. So, $\frac{4}{\sqrt{2}} = 2\sqrt{2}$.

**$$v' = 2\sqrt{2}v$$**

The speed of each of the moving particles after the collision is $2\sqrt{2}v$.

Select Answer:

Visualized Solution

\text{Initial State}

\text{Final State}

\text{Conservation of Linear Momentum}

\text{Momentum along x-axis}

\text{Initial Momentum } (p_{ix})

\text{Total Initial Momentum}

\text{Final Momentum } (p_{fx})

\text{Simplifying Final Momentum}

\text{Equating Momenta}

\text{Solving for } v'

\text{Final Answer}

The Sigma Insight: Conservation of Linear Momentum

The Collision Setup

The Aftermath

The Power of Momentum Conservation

Crunching the Numbers

The Final Result

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