Two particles of masses M and 2M, moving as shown, with speeds of 10 m/s and 5 m/s, collide elastica

Q: Why didn't we use the kinetic energy conservation equation even though it says 'elastic collision'?

In a 2D collision, momentum conservation gives us two independent equations (one for x, one for y). Since we only have two unknowns ($v_1$ and $v_2$), these two equations are sufficient to solve the problem. The kinetic energy equation would be redundant and mathematically more complex to solve.

Q: How do I choose the sign convention for the y-axis?

You can choose either upwards or downwards as positive, as long as you are consistent throughout the equation. In our solution, we took downwards as positive, which made the initial y-momentum of mass $M$ positive and $2M$ negative.

Q: Why is the x-momentum of all particles positive?

Look at the diagram carefully. Both initial particles are moving towards the origin from the left, meaning they are moving to the right (positive x direction). After the collision, both particles move away from the origin to the right. Hence, all x-components are positive.

Q: What if the collision was inelastic?

If the collision was inelastic, momentum would still be conserved in both directions. The equations we derived would still be perfectly valid! The only difference is that kinetic energy would not be conserved.

Q: The calculation with roots looks tedious. Is there a shortcut?

Unfortunately, in this specific problem, you have to plug in the approximate values of $\sqrt{2}$ and $\sqrt{3}$ to find the closest matching option. Always keep the values of $\sqrt{2} \approx 1.414$ and $\sqrt{3} \approx 1.732$ handy for JEE!

Question

Two particles of masses $M$ and $2M$, moving as shown, with speeds of $10	ext{ m/s}$ and $5	ext{ m/s}$, collide elastically at the origin. After the collision, they move along the indicated directions with speed $v_1$ and $v_2$ are nearly

Options:

(A) $6.5	ext{ m/s}$ and $3.2	ext{ m/s}$

(B) $3.2	ext{ m/s}$ and $6.3	ext{ m/s}$

(C) $3.2	ext{ m/s}$ and $12.6	ext{ m/s}$

(D) $6.5	ext{ m/s}$ and $6.3	ext{ m/s}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are watching a cosmic game of billiards. We have two particles, one with mass $M$ and another with mass $2M$, hurtling towards the origin.

The particle of mass $M$ is coming from the top-left at a speed of $10	ext{ m/s}$, making a $30^\circ$ angle with the horizontal. Meanwhile, the heavier particle of mass $2M$ is coming from the bottom-left at $5	ext{ m/s}$, making a $45^\circ$ angle.

They collide at the origin and scatter. The heavier mass $2M$ shoots off to the top-right at an angle of $30^\circ$ with an unknown speed $v_1$. The lighter mass $M$ deflects to the bottom-right at an angle of $45^\circ$ with an unknown speed $v_2$. Our mission is to find these final speeds.

### The Master Equation: Momentum Conservation

Since there are no external forces acting on our system of two particles, we can confidently apply the **Principle of Conservation of Linear Momentum**.

Because this is a two-dimensional collision, we must conserve momentum independently along both the x-axis and the y-axis. This will give us two powerful equations to solve for our two unknowns.

### Conserving Momentum Along the X-Axis

Let's break down the motion horizontally. Both particles are initially moving towards the right, so their initial x-momenta are positive.

For mass $M$, the x-momentum is $M(10\cos 30^\circ)$. For mass $2M$, it is $2M(5\cos 45^\circ)$.

After the collision, both particles continue to move to the right. The final x-momentum for $2M$ is $2M(v_1\cos 30^\circ)$, and for $M$ it is $M(v_2\cos 45^\circ)$.

Equating the initial and final x-momenta, we get:
$$M(10\cos 30^\circ) + 2M(5\cos 45^\circ) = 2M(v_1\cos 30^\circ) + M(v_2\cos 45^\circ)$$

We can immediately cancel out the mass $M$ from every term. Substituting the trigonometric values ($\cos 30^\circ = \frac{\sqrt{3}}{2}$ and $\cos 45^\circ = \frac{1}{\sqrt{2}}$), we simplify the equation to:
$$5\sqrt{3} + 5\sqrt{2} = v_1\sqrt{3} + \frac{v_2}{\sqrt{2}}$$

This is our first master equation.

### Conserving Momentum Along the Y-Axis

Now, let's tackle the vertical motion. Here, we must be very careful with our **sign convention**. Let's choose the downward direction as positive.

Initially, mass $M$ is moving downwards, so its y-momentum is positive: $+M(10\sin 30^\circ)$. Mass $2M$ is moving upwards, so its y-momentum is negative: $-2M(5\sin 45^\circ)$.

After the collision, mass $2M$ moves upwards (negative y-momentum): $-2M(v_1\sin 30^\circ)$. Mass $M$ moves downwards (positive y-momentum): $+M(v_2\sin 45^\circ)$.

Equating the initial and final y-momenta, we get:
$$M(10\sin 30^\circ) - 2M(5\sin 45^\circ) = -2M(v_1\sin 30^\circ) + M(v_2\sin 45^\circ)$$

Again, we cancel $M$ and substitute the sine values ($\sin 30^\circ = \frac{1}{2}$ and $\sin 45^\circ = \frac{1}{\sqrt{2}}$). This simplifies to:
$$5 - 5\sqrt{2} = -v_1 + \frac{v_2}{\sqrt{2}}$$

This is our second master equation.

### Final Calculation: Solving the System

We now have a neat system of two linear equations:
1) $5\sqrt{3} + 5\sqrt{2} = v_1\sqrt{3} + \frac{v_2}{\sqrt{2}}$
2) $5 - 5\sqrt{2} = -v_1 + \frac{v_2}{\sqrt{2}}$

Notice how the term $\frac{v_2}{\sqrt{2}}$ appears in both equations? If we subtract the second equation from the first, this term will beautifully cancel out!

Subtracting (2) from (1):
$$(5\sqrt{3} + 5\sqrt{2}) - (5 - 5\sqrt{2}) = \left(v_1\sqrt{3} + \frac{v_2}{\sqrt{2}}ight) - \left(-v_1 + \frac{v_2}{\sqrt{2}}ight)$$

This simplifies to:
$$5\sqrt{3} + 10\sqrt{2} - 5 = v_1(\sqrt{3} + 1)$$

Now, we isolate $v_1$:
$$v_1 = \frac{5\sqrt{3} + 10\sqrt{2} - 5}{\sqrt{3} + 1}$$

To find the numerical value, we plug in the approximations $\sqrt{3} \approx 1.732$ and $\sqrt{2} \approx 1.414$:
$$v_1 = \frac{5(1.732) + 10(1.414) - 5}{1.732 + 1} = \frac{8.66 + 14.14 - 5}{2.732} = \frac{17.8}{2.732} \approx 6.516 	ext{ m/s}$$

So, $v_1$ is approximately $6.5 	ext{ m/s}$.

To find $v_2$, we substitute this value back into our second equation:
$$\frac{v_2}{\sqrt{2}} = 5 - 5\sqrt{2} + v_1$$
$$\frac{v_2}{\sqrt{2}} = 5 - 7.07 + 6.516 = 4.446$$
$$v_2 = 4.446 	imes \sqrt{2} = 4.446 	imes 1.414 \approx 6.28 	ext{ m/s}$$

So, $v_2$ is approximately $6.3 	ext{ m/s}$.

### The Revelation

Did you notice the trick the examiners played on us? The problem explicitly stated that the particles "collide elastically".

Normally, this means we should also conserve kinetic energy. However, because this was a 2D collision, momentum conservation alone gave us two independent equations. Since we only had two unknowns ($v_1$ and $v_2$), we didn't even need to touch the kinetic energy equation!

Always trust the math, and don't let extra information distract you from the most direct path to the solution.

Animated Solution for Physics - System of Particles: Two particles of masses $M$ and $2 M$ , moving as shown, with speeds of $10 m/s$ and $5 m/s$ , collide elastically at the origin. After the collision, they move along the indicated directions with speed $v_{1}$ and $v_{2}$ are nearly

Select Answer:

Visualized Solution

$Analyzing the Setup$

$Principle of Conservation of Linear Momentum$

$Conserving Momentum Along X-Axis$

$Simplifying the X-Equation$

$Conserving Momentum Along Y-Axis$

$Simplifying the Y-Equation$

$Solving the System of Equations$

$Calculating v_{1}$

$Calculating v_{2}$

$Final Answer$

$The Way Forward$

The Sigma Insight: Oblique Collision

Analyzing the Setup

The Master Equation

Conserving Momentum Along the X-Axis

Conserving Momentum Along the Y-Axis

Final Calculation

The Revelation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Principle of Conservation of Linear Momentum

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conserving Momentum Along X-Axis

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying the X-Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conserving Momentum Along Y-Axis

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying the Y-Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving the System of Equations

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating v1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating v2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward