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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 projectile of mass \(2.0 \mathrm{kg}\) approaches a stationary target body at \(8.0 \mathrm{m} / \mathrm{s} .\) The projectile is deflected through an angle of \(90.0^{\circ}\) and its speed after the collision is $6.0 \mathrm{m} / \mathrm{s}$ What is the speed of the target body after the collision if the collision is perfectly elastic?
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A BMW of mass \(2.0 \times 10^{3} \mathrm{kg}\) is traveling at $42 \mathrm{m} / \mathrm{s} .\( It approaches a \)1.0 \times 10^{3} \mathrm{kg}$ Volkswagen going \(25 \mathrm{m} / \mathrm{s}\) in the same direction and strikes it in the rear. Neither driver applies the brakes. Neglect the relatively small frictional forces on the cars due to the road and due to air resistance. (a) If the collision slows the BMW down to \(33 \mathrm{m} / \mathrm{s},\) what is the speed of the \(\mathrm{VW}\) after the collision? (b) During the collision, which car exerts a larger force on the other, or are the forces equal in magnitude? Explain.
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