An alpha-particle of mass m suffers one-dimensional elastic collision with a nucleus at rest of unkn

Q: Why do we take the velocity of the alpha particle as negative in the momentum equation?

Momentum is a vector quantity. Since the alpha particle is scattered 'directly backwards', its final velocity is in the opposite direction to its initial velocity. If we take the initial direction as positive, the final direction must be negative.

Q: How did we know the nucleus moves forward?

By conservation of momentum, the initial momentum is positive. Since the alpha particle rebounds with a negative momentum, the nucleus must have a positive momentum (move forward) to ensure the total final momentum equals the initial positive momentum.

Q: Could we have used the coefficient of restitution equation instead of energy conservation?

Yes! For an elastic collision, $e=1$, which means the relative velocity of separation equals the relative velocity of approach ($v_2 - (-v_1) = v$). You can use this along with momentum conservation to solve the problem, and it will yield the exact same result.

Q: Why is the remaining kinetic energy of the alpha particle 36%?

The problem states it *loses* 64% of its initial kinetic energy. Therefore, the energy it retains is $100\% - 64\% = 36\%$.

Q: What if the collision wasn't elastic?

If the collision were inelastic, kinetic energy would not be conserved. Some energy would be lost to heat or deformation, and we wouldn't be able to equate the lost energy of the alpha particle to the gained energy of the nucleus.

Question

An $\alpha$-particle of mass $m$ suffers one-dimensional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing 64% of its initial kinetic energy. The mass of the nucleus is

Options:

(A) 1.5 m

(B) 4 m

(C) 3.5 m

(D) 2 m

Sigma Percentile · Accepted Answer

Imagine you are observing a microscopic game of billiards. An $\alpha$-particle, a tiny but highly energetic projectile of mass $m$, is cruising along with a velocity $v$. Up ahead, a mysterious nucleus of unknown mass $M$ sits perfectly at rest, waiting for the inevitable impact.

This is a classic one-dimensional head-on collision. But this isn't just any collision; it's an **elastic collision**, which means the universe is keeping a strict, perfect accounting of both momentum and kinetic energy. No energy is lost to heat or deformation.

### Tracking the Energy Flow

The problem gives us a fascinating clue: after the collision, the $\alpha$-particle is scattered directly backwards, losing $64\%$ of its initial kinetic energy.

What does it mean to lose $64\%$? It means the $\alpha$-particle retains exactly $36\%$ of its original energy. Let's translate this into the language of mathematics. If $v_1$ is the rebound speed of the $\alpha$-particle, we can write:

$$\frac{1}{2} m v_1^2 = 0.36 \left( \frac{1}{2} m v^2 \right)$$

Taking the square root of both sides beautifully simplifies this to:

$$v_1 = 0.6 v$$

But wait, where did that lost $64\%$ of energy go? Because the collision is perfectly elastic, that energy didn't just vanish; it was entirely transferred to the stationary nucleus! The nucleus wakes up and surges forward with a new speed, $v_2$. We can equate its new kinetic energy to the energy lost by the $\alpha$-particle:

$$\frac{1}{2} M v_2^2 = 0.64 \left( \frac{1}{2} m v^2 \right)$$

Solving for $v_2$, we get:

$$v_2 = \sqrt{\frac{m}{M}} (0.8 v)$$

### The Momentum Master Equation

Now we have the speeds of both particles after the collision. To find the unknown mass $M$, we need to bring in the heavy hitter of mechanics: **Conservation of Linear Momentum**.

Unlike kinetic energy, momentum is a vector. It cares deeply about direction. Initially, only the $\alpha$-particle is moving, so the total initial momentum is simply $mv$ (taking the forward direction as positive).

After the collision, the nucleus moves forward with momentum $M v_2$. However, the $\alpha$-particle has rebounded; it is moving *backwards*. Therefore, its momentum is negative, specifically $-m v_1$. This minus sign is the most common trap in collision problems!

Equating the initial and final momentum:

$$m v = M v_2 - m v_1$$

### The Final Reveal

Let's plug in the expressions for $v_1$ and $v_2$ that we derived from our energy analysis:

$$m v = M \left( \sqrt{\frac{m}{M}} (0.8 v) \right) - m (0.6 v)$$

Notice that we have an $m(0.6v)$ term on the right. Let's move it to the left side to group the terms containing $m$:

$$m v + 0.6 m v = \sqrt{m M} (0.8 v)$$

$$1.6 m v = \sqrt{m M} (0.8 v)$$

The velocity $v$ cancels out entirely, leaving us with a pure mass relationship. Dividing both sides by $0.8$:

$$2 m = \sqrt{m M}$$

To get rid of the square root, we simply square both sides:

$$4 m^2 = m M$$

Dividing by $m$, we arrive at our grand conclusion:

$$M = 4m$$

The mass of the unknown nucleus is exactly four times the mass of the $\alpha$-particle. Physically, this makes perfect sense. For a lighter particle to rebound with such a significant fraction of its speed, it must have struck something substantially heavier—like a tennis ball bouncing off a bowling ball. The math and the physics align perfectly!

Animated Solution for Physics - System of Particles: An $α$ -particle of mass $m$ suffers one-dimensional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing 64% of its initial kinetic energy. The mass of the nucleus is

Select Answer:

Visualized Solution

\text{Visualizing the Collision}

\text{After the Collision}

\text{Kinetic Energy of } \alpha \text{-particle}

\text{Kinetic Energy of Nucleus}

\text{Conservation of Linear Momentum}

\text{Solving for } M

\text{Conclusion}

The Sigma Insight: Head-on Collision

Tracking the Energy Flow

The Momentum Master Equation

The Final Reveal

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