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Chapter 7: Problem 59

An alpha particle collides with an oxygen nucleus that is initially at rest. The alpha particle is scattered at an angle of \(64.0^{\circ}\) from its initial direction of motion, and the oxygen nucleus recoils at an angle of \(51.0^{\circ}\) on the opposite side of that initial direction. The final speed of the nucleus is \(1.20 \times 10^{5} \mathrm{~m} / \mathrm{s}\). Find (a) the final speed and (b) the initial speed of the alpha particle. (In atomic mass units, the mass of an alpha particle is \(4.0 \mathrm{u}\), and the mass of an oxygen nucleus is \(16 \mathrm{u}\).)

Short Answer

Expert verified
Final speed of alpha particle: 4.15 脳 10鈦 m/s; Initial speed of alpha particle: 6.41 脳 10鈦 m/s

Step by step solution

01

- Understand the Collision

An alpha particle collides with an oxygen nucleus. The alpha particle has mass 4u and the oxygen nucleus has mass 16u. Post-collision, the alpha particle is scattered at an angle of 64.0掳 from its original path, and the oxygen nucleus recoils at an angle of 51.0掳 on the opposite side.
02

- Set Up Conservation of Momentum

Use the principle of conservation of momentum. Let's resolve the momentum into two components: x (original direction of the alpha particle) and y (perpendicular to that). Define the initial speed of the alpha particle as v岬 and its final speed as v鈧. Define the final speed of the oxygen nucleus as v鈧 = 1.20 脳 10鈦 m/s.
03

- Momentum Equation in the x-direction

In the x-direction: defined as 4u * v岬 = 4u * v鈧 * cos(64.0掳) + 16u * v鈧 * cos(51.0掳)
04

- Momentum Equation in the y-direction

In the y-direction: defined as 0 = 4u * v鈧 * sin(64.0掳) - 16u * v鈧 * sin(51.0掳)
05

- Solve for v鈧

From the y-component equation: 4 * v鈧 * sin(64.0掳) = 16 * v鈧 * sin(51.0掳)Substituting for v鈧: distribute and divide by trigonometric functions and masses v鈧 * sin(64.0掳) = 4 * 1.20 脳 10鈦 * sin(51.0掳) v鈧 = (4 * 1.20 脳 10鈦 * sin(51.0掳)) / sin(64.0掳)
06

- Calculate the Final Speed of alpha particle

Substituting values and simplifying: sin(51.0掳) 鈮 0.777, and sin(64.0掳) 鈮 0.899 therefore, v鈧 = (4 * 1.20 脳 10鈦 m/s * 0.777) / 0.899 v鈧 鈮 4.15 脳 10鈦 m/s
07

- Solve for v岬

Using x-component equation and substituting v鈧 found before: 4 * v岬 = 4 * 4.15 脳 10鈦 * cos(64.0掳) + 16 * 1.20 脳 10鈦 * cos(51.0掳) v岬 = v鈧 * cos(64.0掳) + 4 * v鈧 * cos(51.0掳)cos(64.0掳) 鈮 0.438, and cos(51.0掳) 鈮 0.629 v岬 = 4.15 脳 10鈦 * 0.438 + 4 * 1.20 脳 10鈦 * 0.629 v岬 鈮 6.41 脳 10鈦 m/s

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

collision
A collision occurs when two particles come into contact, exchanging energy and momentum. In our scenario, an alpha particle collides with an oxygen nucleus. The alpha particle has a mass of 4u, while the oxygen nucleus has a heavier mass of 16u. The angles at which they scatter and recoil provide crucial information. The alpha particle scatters at an angle of 64.0掳 from its initial path. Meanwhile, the oxygen nucleus recoils at an angle of 51.0掳 on the opposite side. Analyzing such collisions helps us understand the principles behind momentum conservation and energy exchange during the interaction.
momentum components
In physics, momentum is conserved in collisions, meaning the total momentum before and after the collision remains constant. To apply this, we resolve the momentum into two components:
  • x-direction (original direction of the alpha particle)
  • y-direction (perpendicular to that)
By defining initial and final speeds, we set up equations using conservation principles. The conservation of momentum in each direction helps break down complex problems into solvable parts. For example, the x-direction equation: 4u * v岬 = 4u * v鈧 * cos(64.0掳) + 16u * v鈧 * cos(51.0掳); and the y-direction equation: 0 = 4u * v鈧 * sin(64.0掳) - 16u * v鈧 * sin(51.0掳), are pivotal in solving for the unknown speeds.
final speed calculation
Determining the final speeds of particles after a collision involves using trigonometric functions and momentum equations. We start with the y-direction equation, because it only involves final speeds and known angles. After isolating v鈧 (the final speed of the alpha particle): v鈧 * sin(64.0掳) = 4 * 1.20 脳 10鈦 * sin(51.0掳), we solve by substituting known trigonometric values. With sin(51.0掳) 鈮 0.777 and sin(64.0掳) 鈮 0.899, we find v鈧 鈮 4.15 脳 10鈦 m/s. This process involves substituting values, simplifying, and solving the equations step-by-step to get the final speeds accurately.
initial speed calculation
To find the initial speed v岬 of the alpha particle, we use the x-direction momentum equation. Substituting the final speed v鈧 found earlier into the equation: 4 * v岬 = 4 * 4.15 脳 10鈦 * cos(64.0掳) + 16 * 1.20 脳 10鈦 * cos(51.0掳), and using trigonometric values cos(64.0掳) 鈮 0.438 and cos(51.0掳) 鈮 0.629, we solve for v岬. Thus, v岬 = 4.15 脳 10鈦 * 0.438 + 4 * 1.20 脳 10鈦 * 0.629, leading to v岬 鈮 6.41 脳 10鈦 m/s. This approach ensures that the total momentum is conserved, adhering to the foundational principles of physics.

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Most popular questions from this chapter

A \(0.70 \mathrm{~kg}\) ball is moving horizontally with a speed of \(5.0 \mathrm{~m} / \mathrm{s}\) when it strikes a vertical wall. The ball rebounds with a speed of \(2.0 \mathrm{~m} / \mathrm{s}\). What is the magnitude of the change in translational momentum of the ball?

\(\mathrm{A}\) barge with mass \(1.50 \times\) \(10^{5} \mathrm{~kg}\) is proceeding downriver at \(6.2 \mathrm{~m} / \mathrm{s}\) in heavy fog when it collides with a barge heading directly across the river (see Fig. 7-31). The second barge has mass \(2.78\) \(\times 10^{5} \mathrm{~kg}\) and before the collision is moving at \(4.3\) \(\mathrm{m} / \mathrm{s} . \quad\) Immediately after impact, the second barge finds its course deflected by \(18^{\circ}\) in the downriver direction and its speed increased to \(5.1 \mathrm{~m} / \mathrm{s}\). The river current is approximately zero at the time of the accident. What are the speed and direction of motion of the first barge immediately after the collision?

A professor of physics is going ice skating for the first time. He has gotten himself into the middle of an ice rink and cannot figure out how to make the skates work. Every motion he makes simply causes his feet to slip on the ice and leaves him in the same place he started. He decides that he can get off the ice by throwing his gloves in the opposite direction. (a) Suppose he has a mass \(M\) and his gloves have a mass \(m\). If he throws the gloves as hard as he can away from him, they leave his hand with a velocity \(\vec{v}_{\text {glove }}\) Explain whether or not he will move. If he does move, calculate his velocity, \(\vec{v}_{\text {prof }}\) (b) Discuss his motion from the point of view of the forces acting on him. (c) If the ice rink is \(10 \mathrm{~m}\) in diameter and the skater starts in the center, estimate how long it will take him to reach the edge, assuming there is no friction at all.

A bullet of mass \(4.5 \mathrm{~g}\) is fired horizontally into a \(2.4 \mathrm{~kg}\) wooden block at rest on a horizontal surface. The bullet is embedded in the block. The speed of the block immediately after the bullet stops relative to it is \(2.7 \mathrm{~m} / \mathrm{s}\). At what speed is the bullet fired?

A \(4.0 \mathrm{~kg}\) mess kit sliding on a frictionless surface explodes into two \(2.0 \mathrm{~kg}\) parts, one moving at \(3.0 \mathrm{~m} / \mathrm{s}\), due north, and the other at \(5.0 \mathrm{~m} / \mathrm{s}, 30^{\circ}\) north of east. What is the original speed of the mess kit?
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