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Chapter 7: Problem 39

A helium atom (mass \(4.00 \mathrm{u}\) ) moving at $618 \mathrm{m} / \mathrm{s}$ to the right collides with an oxygen molecule (mass 32.0 u) moving in the same direction at \(412 \mathrm{m} / \mathrm{s}\). After the collision, the oxygen molecule moves at \(456 \mathrm{m} / \mathrm{s}\) to the right. What is the velocity of the helium atom after the collision?

Short Answer

Expert verified
Answer: The final velocity of the helium atom after the collision is 266 m/s to the right.

Step by step solution

01

Write the conservation of momentum formula

We will use the conservation of momentum principle, which states that the total momentum of a closed system is conserved during a collision. Mathematically, the principle can be written as: $$m_1v_1_initial + m_2v_2_initial = m_1v_1_final + m_2v_2_final$$
02

Plug in given values into the formula

We are given the following values: $$m_1 (helium) = 4.00\,\text{u}$$ $$v_1_initial (helium) = 618\,\text{m/s}$$ $$m_2 (oxygen) = 32.0\,\text{u}$$ $$v_2_initial (oxygen) = 412\,\text{m/s}$$ $$v_2_final (oxygen) = 456\,\text{m/s}$$ Now, we need to find the final velocity of helium atom (\(v_1_final\)). Substitute the given values into the conservation of momentum formula: $$4.00\,\text{u} \times 618\,\text{m/s} + 32.0\,\text{u} \times 412\,\text{m/s} = 4.00\,\text{u} \times v_1_final + 32.0\,\text{u} \times 456\,\text{m/s}$$
03

Solve for the final velocity of helium atom \(v_1_final\)

Rearrange the equation to isolate \(v_1_final\): $$v_1_final = \frac{(4.00\,\text{u} \times 618\,\text{m/s} + 32.0\,\text{u} \times 412\,\text{m/s}) - (32.0\,\text{u} \times 456\,\text{m/s})}{4.00\,\text{u}}$$ Now, perform the calculations: $$v_1_final = \frac{(4.00 \times 618 + 32.0 \times 412) - (32.0 \times 456)}{4.00}$$ $$v_1_final = \frac{2472 + 13184 - 14592}{4}$$ $$v_1_final = \frac{1064}{4}$$ $$v_1_final = 266\,\text{m/s}$$ So, the final velocity of the helium atom after the collision is \(266\,\text{m/s}\) to the right.

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Most popular questions from this chapter

A 0.15-kg baseball is pitched with a speed of \(35 \mathrm{m} / \mathrm{s}\) \((78 \mathrm{mph}) .\) When the ball hits the catcher's glove, the glove moves back by \(5.0 \mathrm{cm}(2 \text { in. })\) as it stops the ball. (a) What was the change in momentum of the baseball? (b) What impulse was applied to the baseball? (c) Assuming a constant acceleration of the ball, what was the average force applied by the catcher's glove?
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