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Chapter 7: Problem 27

Particle A is at the origin and has a mass of \(30.0 \mathrm{g} .\) Particle B has a mass of \(10.0 \mathrm{g} .\) Where must particle \(\mathrm{B}\) be located if the coordinates of the \(\mathrm{cm}\) are \((x, y)=(2.0 \mathrm{cm},\) $5.0 \mathrm{cm}) ?$

Short Answer

Expert verified
Answer: The coordinates of Particle B are (8.0 cm, 20.0 cm).

Step by step solution

01

Write down the formula for the center of mass in two dimensions

The formula for the center of mass in two dimensions is given by: \(x_{cm} = \frac{m_Ax_A + m_Bx_B}{m_A + m_B}\) \(y_{cm} = \frac{m_Ay_A + m_By_B}{m_A + m_B}\) where \(m_A\) and \(m_B\) are the masses of particles A and B, respectively, and \((x_A, y_A)\) and \((x_B, y_B)\) are the coordinates of particles A and B, respectively.
02

Substitute the given values into the x-coordinate formula for center of mass

We are given \(x_{cm} = 2.0 cm\), \(m_A = 30.0 g\), \(x_A = 0\), and \(m_B = 10.0 g\). Plugging these values into the x-coordinate formula, we get: \(2.0 = \frac{30.0 \times 0 + 10.0 \times x_B}{30.0 + 10.0}\)
03

Solve for the x-coordinate of Particle B

Simplifying and solving for \(x_B\), we obtain: \(2.0 = \frac{10.0x_B}{40.0}\) \(x_B = \frac{2.0 \times 40.0}{10.0}\) \(x_B = 8.0 \;\mathrm{cm}\)
04

Substitute the given values into the y-coordinate formula for center of mass

We are given \(y_{cm} = 5.0 cm\), \(m_A = 30.0 g\), \(y_A = 0\), and \(m_B = 10.0 g\). Plugging these values into the y-coordinate formula, we get: \(5.0 = \frac{30.0 \times 0 + 10.0 \times y_B}{30.0 + 10.0}\)
05

Solve for the y-coordinate of Particle B

Simplifying and solving for \(y_B\), we obtain: \(5.0 = \frac{10.0y_B}{40.0}\) \(y_B = \frac{5.0 \times 40.0}{10.0}\) \(y_B = 20.0 \;\mathrm{cm}\)
06

Combine x and y coordinates for Particle B

Now that we have found both the x and y coordinates of Particle B, we can combine them to find its location: The location of Particle B is \((8.0 \;\mathrm{cm}, 20.0 \;\mathrm{cm})\).

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Most popular questions from this chapter

A rifle has a mass of \(4.5 \mathrm{kg}\) and it fires a bullet of mass $10.0 \mathrm{g}\( at a muzzle speed of \)820 \mathrm{m} / \mathrm{s} .$ What is the recoil speed of the rifle as the bullet leaves the gun barrel?
A 2.0-kg block is moving to the right at \(1.0 \mathrm{m} / \mathrm{s}\) just before it strikes and sticks to a 1.0-kg block initially at rest. What is the total momentum of the two blocks after the collision?
A 75-kg man is at rest on ice skates. A 0.20-kg ball is thrown to him. The ball is moving horizontally at \(25 \mathrm{m} / \mathrm{s}\) just before the man catches it. How fast is the man moving just after he catches the ball?
A 6.0 -kg object is at rest on a perfectly frictionless surface when it is struck head-on by a 2.0 -kg object moving at \(10 \mathrm{m} / \mathrm{s} .\) If the collision is perfectly elastic, what is the speed of the 6.0-kg object after the collision? [Hint: You will need two equations.]
A 3.0-kg body is initially moving northward at \(15 \mathrm{m} / \mathrm{s}\) Then a force of \(15 \mathrm{N},\) toward the east, acts on it for 4.0 s. (a) At the end of the 4.0 s, what is the body's final velocity? (b) What is the change in momentum during the \(4.0 \mathrm{s} ?\)
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