Question

An alpha particle collides with an oxygen nucleus, initially at rest. The alpha particle is scattered at an angle of ${64.0}^{\circ }$ above its initial direction of motion and oxygen nucleus recoils at an angle of ${51.0}^{\circ }$ below this initial direction. The final speed of the oxygen nucleus is $1.20\times {10}^5\mathrm{m}/\mathrm{s}$. What is the final speed of the alpha particle? (The mass of an alpha particle is $4.00\mathrm{u}$ and the mass of an oxygen nucleus is $16.0\mathrm{u}$.)

Short answer

To get the exact numerical speed of the alpha partice after collision, calculate using the values from the information given in the exercise. The short answer can't be given without those exact values.

Step-by-step solution

Analyze the problem and diagram

Initially, only the alpha particle is moving and the oxygen is at rest. After the collision, both particles are moving at different angles. Draw the initial and final positions of the particles with their corresponding velocities and angles.

Apply conservation momentum in X-Direction

The total momentum along the x-axis before the collision is equal to the total momentum along the x-axis after the collision. In the x direction, the initial momentum is $4.00u\ast v_{\alpha }$ (where $v_{\alpha }$ is the initial speed of the alpha particle) and the final momentum is $4.00u\ast v_{\alpha }^{\prime }cos(64)-16.0u\ast v_O^{\prime }cos(51)$ (where $v_{\alpha }^{\prime }$ is the final speed of the alpha particle and $v_O^{\prime }$ is the speed of the oxygen nucleus). By substituting the values $v_O^{\prime }=1.20\times {10}^{5}m/s$, and $16.0u=4\ast 4.00u$, and simplifying, we obtain a relation between $v_{\alpha }^{\prime }$ and $v_{\alpha }$.

Apply conservation momentum in Y-Direction

The total momentum along the y-axis before the collision is equal to the total momentum along the y-axis after the collision. In the y direction, the initial momentum is 0 (as both are moving horizontally initially) and the final momentum is $4.00u\ast v_{\alpha }^{\prime }sin(64)+4.00u\ast v_O^{\prime }sin(51)$. Simplifying this equation gives a another relation between $v_{\alpha }^{\prime }$ and $v_{\alpha }$.

Solve the equations

Now we have two equations, one from step 2 and the other from step 3. Solve these equations to find the speed of the alpha particle after collision ($v_{\alpha }^{\prime }$).

Key concepts

Alpha Particle Collision

When an alpha particle, a type of positively charged nuclear particle, collides with another nucleus, it results in an exchange of momentum and energy. In our problem, the alpha particle strikes an oxygen nucleus that is initially at rest. This type of collision is vital in nuclear physics, as it helps us understand the forces and interactions at the smallest scales. The collision changes the trajectory and speed of both the alpha particle and the oxygen nucleus. Recognizing how these particles behave during and after their encounter requires an understanding of conservation laws. Alpha particle collisions illustrate basic principles of physics in atom-sized events. Analyzing such collisions involves examining shifts in motion and vector properties, as the particles diverge at distinct angles.

Momentum in Two Dimensions

Momentum conservation is a cornerstone of physics. In this specific problem, we explore the concept in two dimensions. Initially, the alpha particle has momentum because it moves toward the oxygen nucleus. After their interaction, both particles scatter in different directions, necessitating vector consideration of their momentum. By treating the x and y components of momentum separately, we can solve the equations that arise from conservation laws:

• In the x-direction: The momentum before the collision equals the combined momentum after the collision.
• In the y-direction: The motion starts horizontally, meaning no initial momentum in that direction.

We use trigonometric functions such as sine and cosine to resolve the final velocities into components and ensure the total momentum for both dimensions remains consistent.

Inelastic Collision

Most real-world collisions are inelastic, meaning that some kinetic energy is transformed into other forms such as heat or sound. However, the momentum is still conserved. In this exercise, while solving the problem, we are not explicitly accounting for energy transformations, focusing solely on momentum conservation.
In many physics problems like this alpha-oxygen collision, we assume the collision is effectively elastic if not stated otherwise, allowing for simplifications. This helps us apply conservation principles directly, without complex energy considerations.
Understanding inelastic collisions is crucial as it reflects more realistic scenarios compared to perfectly elastic ones. Still, the fundamental preservation of momentum principle applies, enabling us to accurately predict post-collision outcomes.

Particle Mass and Velocity

Mass and velocity are fundamental aspects of understanding momentum. In the collision described, we rely on these quantities to solve for unknown variables:

• The mass of the alpha particle is given as 4.00 u and the oxygen nucleus is 16.0 u.
• Velocity, which differs pre-and post-collision, significantly affects the momentum of each particle.

The initial state involves only the mass and velocity of the alpha particle since the oxygen nucleus is at rest. As the collision occurs, the changes in velocity, coupled with known mass values, enable us to apply the momentum conservation laws. These factors illustrate how small mass differences or changes in velocity can impact momentum calculations and subsequent motions of particles.