A ball with a speed of 9 m/s collides with another identical ball at rest. After the collision, the

Q: Why didn't we use the $9\text{ m/s}$ initial speed?

Because we only needed the ratio of the final velocities! By analyzing the y-direction where the initial momentum is zero, the initial speed becomes irrelevant. It's a neat trick to save time.

Q: Does this mean the collision is elastic?

Not necessarily. We only used momentum conservation, which holds true for both elastic and inelastic collisions. To check if it's elastic, we would need to calculate and compare the initial and final kinetic energies.

Q: What if the masses were different?

If the masses were $m_1$ and $m_2$, the y-direction equation would be $m_1 v_1 \sin 30^\circ = m_2 v_2 \sin 30^\circ$. The ratio $v_1/v_2$ would then depend on the inverse ratio of their masses, $m_2/m_1$.

Q: Why is the final momentum in the y-direction subtracted?

Momentum is a vector. If we take upwards as positive, the first ball has a positive y-momentum ($+m v_1 \sin 30^\circ$), and the second ball, moving downwards, has a negative y-momentum ($-m v_2 \sin 30^\circ$). Their sum must be zero.

Q: Can we solve this using the x-direction momentum?

Yes, but it's more complicated. The x-direction equation is $m(9) = m v_1 \cos 30^\circ + m v_2 \cos 30^\circ$. You would need to use this along with the y-direction equation to find the exact values of $v_1$ and $v_2$, but for just the ratio, the y-direction alone is sufficient and much faster.

Question

A ball with a speed of $9	ext{ m/s}$ collides with another identical ball at rest. After the collision, the direction of each ball makes an angle of $30^\circ$ with the original direction. The ratio of velocities of the balls after collision is $x : y$, where $x$ is ........... .

Sigma Percentile · Accepted Answer

### The Setup: Visualizing the Collision

Imagine a cosmic game of billiards. A ball of mass $m$ is zooming along the x-axis at a brisk $9	ext{ m/s}$. It is on a direct collision course with another identical ball that is completely at rest.

Boom! The collision happens. But instead of moving straight ahead, the balls scatter. They both deflect at an angle of $30^\circ$ from the original path, one shooting upwards and the other downwards.

We are tasked with finding the ratio of their final speeds, $v_1$ and $v_2$. At first glance, you might think we need to set up complex equations involving the $9	ext{ m/s}$ initial speed and kinetic energy. But there is a much more elegant way to solve this.

### The Secret Weapon: Momentum Conservation

In physics, when no external forces are acting on a system, momentum is always conserved. This is a fundamental law of nature. Because momentum is a vector, it is conserved in every direction independently.

This means we can look at the x-direction and the y-direction as two separate problems. The initial motion is entirely horizontal (along the x-axis). Therefore, the initial momentum in the vertical (y-direction) is exactly zero.

### The Y-Direction Trick

Let's focus solely on the y-direction. Before the collision, nothing is moving up or down. So, $p_{iy} = 0$.

After the collision, the first ball has an upward velocity component of $v_1 \sin 30^\circ$. Its vertical momentum is $+m v_1 \sin 30^\circ$. The second ball has a downward velocity component of $v_2 \sin 30^\circ$, giving it a vertical momentum of $-m v_2 \sin 30^\circ$.

According to the law of conservation of momentum, the total final vertical momentum must equal the initial vertical momentum.

$$0 = m v_1 \sin 30^\circ - m v_2 \sin 30^\circ$$

### The Grand Finale: Finding the Ratio

Now, we just need to do a little bit of algebra. We can move the negative term to the other side of the equation.

$$m v_1 \sin 30^\circ = m v_2 \sin 30^\circ$$

Notice how beautifully things cancel out! The mass $m$ is the same for both balls, so it disappears. The term $\sin 30^\circ$ (which is $1/2$) is also on both sides, so it cancels out as well.

$$v_1 = v_2$$

This tells us that both balls fly off with the exact same speed! Therefore, the ratio of their velocities $v_1 : v_2$ is simply $1 : 1$.

The problem states that this ratio is $x : y$. By comparing our result, it is crystal clear that **$x = 1$**. We solved the entire problem without ever needing to use the $9	ext{ m/s}$ initial speed!

Enter Numerical Value:

Visualized Solution

$v_{ini t ia l} = 9 m/s$

$θ = 3 0^{\circ}$

$p_{i y} = p_{f y}$

$0 = m v_{1} sin 3 0^{\circ} - m v_{2} sin 3 0^{\circ}$

$0 = m (\frac{1}{2}) v_{1} - m (\frac{1}{2}) v_{2}$

$\frac{v _{1}}{v _{2}} = \frac{1}{1}$

$x = 1$

The Sigma Insight: Oblique Collision

The Setup

The Secret Weapon

The Y-Direction Trick

The Grand Finale

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vinitial​=9 m/s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ=30∘

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 0=mv1​sin30∘−mv2​sin30∘

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 0=m(21​)v1​−m(21​)v2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v2​v1​​=11​

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