A particle of mass m is moving along the X-axis with initial velocity u i^. It collides elastically

Q: Why do we only conserve momentum in the Y-direction?

We can conserve it in the X-direction too! However, the Y-direction is much easier because the initial momentum is zero, and it directly involves the sines of the angles given in the problem's condition.

Q: Why is kinetic energy conserved in this problem?

The problem explicitly states that the collision is 'elastic'. In physics, a perfectly elastic collision is defined as one where the total kinetic energy of the system is conserved.

Q: What happens to the X-momentum? Is it not conserved?

It is absolutely conserved! The equation would be $mu = mv_1\cos\theta_1 + 10mv_2\cos\theta_2$. However, we don't need to use this equation to solve for $n$ because the energy equation provides the final necessary link.

Q: Why does the small mass lose exactly half its energy? Is this a rule?

No, it's not a general rule. It is a specific condition given for this particular collision scenario to help us solve the problem.

Q: Can the value of $n$ be negative?

No. Since $n$ appears inside a square root in the problem statement ($\sqrt{n}$), it must be a non-negative real number for the expression to be physically meaningful.

Question

A particle of mass $m$ is moving along the X-axis with initial velocity $u\hat{i}$. It collides elastically with a particle of mass $10 m$ at rest and then moves with half its initial kinetic energy (see figure). If $\sin	heta_1 = \sqrt{n}\sin	heta_2$, then value of $n$ is ........

Sigma Percentile · Accepted Answer

### The Elastic Dance of Particles: Solving a 2D Collision

Imagine a small particle of mass $m$ zooming along the X-axis with velocity $u$, heading straight for a massive particle of $10m$ sitting at rest. They collide elastically, scattering in different directions. This is a classic 2D collision problem that requires us to carefully balance momentum and energy.

### The Kinetic Energy Clue

The problem states a very specific condition: the small particle moves away with exactly half of its initial kinetic energy. This is our first major clue. We can write this mathematically as:

$$K_{f1} = \frac{1}{2} K_{i1}$$

$$\frac{1}{2} m v_1^2 = \frac{1}{2} \left(\frac{1}{2} m u^2ight)$$

This beautifully simplifies to give us its final speed, $v_1 = \frac{u}{\sqrt{2}}$.

### Balancing the Y-Momentum

Now, since there are no external forces acting on the system, **linear momentum is conserved**. Let's look at the Y-direction. Initially, there was absolutely no vertical motion. Therefore, the upward momentum of the small mass must perfectly cancel the downward momentum of the large mass after the collision.

$$p_{yi} = p_{yf}$$

$$0 = m v_1 \sin	heta_1 - 10m v_2 \sin	heta_2$$

$$\Rightarrow m v_1 \sin	heta_1 = 10m v_2 \sin	heta_2$$

### Utilizing the Angle Relation

The problem gives us a neat relation between the scattering angles: $\sin	heta_1 = \sqrt{n}\sin	heta_2$. Let's substitute this, along with our expression for $v_1$, into the momentum equation we just derived.

$$m \left(\frac{u}{\sqrt{2}}ight) (\sqrt{n}\sin	heta_2) = 10m v_2 \sin	heta_2$$

The $\sin	heta_2$ terms elegantly cancel out, leaving us with an expression for $v_2$ entirely in terms of $u$ and $n$:

$$v_2 = \frac{u\sqrt{n}}{10\sqrt{2}}$$

### The Grand Finale: Elasticity

Because the collision is perfectly **elastic**, the total kinetic energy before the crash must equal the total kinetic energy after. The initial energy of the small mass is now split between the two masses.

$$K_i = K_f$$

$$\frac{1}{2} m u^2 = \frac{1}{2} m v_1^2 + \frac{1}{2} (10m) v_2^2$$

Let's plug in our expressions for $v_1$ and $v_2$:

$$\frac{1}{2} m u^2 = \frac{1}{2} m \left(\frac{u}{\sqrt{2}}ight)^2 + 5m \left(\frac{u\sqrt{n}}{10\sqrt{2}}ight)^2$$

Notice how every term contains an $m$ and a $u^2$. We can happily cancel them out!

$$1 = \frac{1}{2} + 10 \left(\frac{n}{200}ight)$$

$$\frac{1}{2} = \frac{n}{20}$$

Solving this simple fraction gives us our final answer:

$$n = 10$$

Enter Numerical Value:

Visualized Solution

Visualizing the Collision

Final Velocity of Mass $m$

Momentum Conservation in Y-direction

Substituting the Angle Relation

Conservation of Kinetic Energy

Solving for $n$

The Sigma Insight: Oblique Collision

The Elastic Dance of Particles

The Kinetic Energy Clue

Balancing the Y-Momentum

Utilizing the Angle Relation

The Grand Finale

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