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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

Two cars \(A\) and \(B\) slide on an icy road as they attempt to stop at a traffic light. The mass of \(A\) is \(1100 \mathrm{~kg}\), and the mass of \(B\) is \(1400 \mathrm{~kg}\). The coefficient of kinetic friction between the locked wheels of either car and the road is 0.13. Car \(A\) succeeds in stopping at the light, but car \(B\) cannot stop and rear-ends car \(A\). After the collision, \(A\) stops \(8.2 \mathrm{~m}\) ahead of its position at impact, and \(B 6.1 \mathrm{~m}\) ahead; see Fig. 7-27. Both drivers had their brakes locked throughout the incident. Using the material in Chapters 2 and 6 , find the speed of (a) car \(A\) and (b) car \(B\) immediately after impact. (c) Use conservation of translational momentum to find the speed at which car \(B\) struck car \(A\). On what grounds can the use of momentum conservation be criticized here?

In February 1955, a paratrooper fell \(370 \mathrm{~m}\) from an airplane without being able to open his chute but happened to land in snow, suffering only minor injuries. Assume that his speed at impact was \(56 \mathrm{~m} / \mathrm{s}\) (terminal speed), that his mass (including gear) was \(85 \mathrm{~kg}\), and that the magnitude of the force on him from the snow was at the survivable limit of \(1.2 \times 10^{5} \mathrm{~N}\). What are (a) the minimum depth of snow that would have stopped him safely and (b) the magnitude of the impulse on him from the snow?

In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball moves at \(3.50 \mathrm{~m} / \mathrm{s}\) along a line making an angle of \(22.0^{\circ}\) with its original direction of motion, and the second ball has a speed of \(2.00 \mathrm{~m} / \mathrm{s}\). Find (a) the angle between the direction of motion of the second ball and the original direction of motion of the cue ball and (b) the original speed of the cue ball.

It is well known that bullets and other missiles fired at Superman simply bounce off his chest (Fig.7-22). Suppose that a gangster sprays Superman's chest with \(3 \mathrm{~g}\) bullets at the rate of 100 bullets/min, and the speed of each bullet is \(500 \mathrm{~m} / \mathrm{s}\). Suppose too that the bullets rebound straight back with no change in speed. What is the magnitude of the average force on Superman's chest from the stream of bullets?

The last stage of a rocket, which is traveling at a speed of \(7600 \mathrm{~m} / \mathrm{s}\), consists of two parts that are clamped together: a rocket case with a mass of \(290.0 \mathrm{~kg}\) and a payload capsule with a mass of \(150.0 \mathrm{~kg}\). When the clamp is released, a compressed spring causes the two parts to separate with a relative speed of \(910.0 \mathrm{~m} / \mathrm{s}\). What are the speeds of (a) the rocket case and (b) the payload after they have separated? Assume that all velocities are along the same line.
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