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Chapter 7: Problem 34

Consider two falling bodies. Their masses are \(3.0 \mathrm{kg}\) and $4.0 \mathrm{kg} .\( At time \)t=0,$ the two are released from rest. What is the velocity of their \(\mathrm{CM}\) at \(t=10.0 \mathrm{s} ?\) Ignore air resistance.

Short Answer

Expert verified
Answer: The velocity of the center of mass for the two falling bodies at 10.0 seconds is 98.1 m/s.

Step by step solution

01

Identify the equation of motion for a falling object

Since both bodies are falling freely under gravity and ignoring air resistance, we can find their individual velocities after 10 seconds using one of the equations of motion: \(v = u + gt\) Here, \(v\) is the final velocity, \(u\) is the initial velocity (0 m/s), \(g\) is the acceleration due to gravity (\(9.81\,\text{m/s}^2\)), and \(t\) is the time (10 seconds).
02

Calculate individual velocities at 10 seconds

For the 3 kg mass (mass 1): \(v_1 = u + gt_1 = 0 + (9.81)(10) = 98.1\,\text{m/s}\) For the 4 kg mass (mass 2): \(v_2 = u + gt_2 = 0 + (9.81)(10) = 98.1\,\text{m/s}\) Notice that both velocities are equal because the motion is under constant gravitational force, and air resistance is ignored.
03

Determine the equation for the center of mass velocity

The velocity of the center of mass (CM) for a system of two particles is given by: \(V_{CM} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2}\) Here, \(V_{CM}\) is the velocity of the center of mass, \(m_1\) and \(m_2\) are the masses of the two particles, and \(v_1\) and \(v_2\) are their respective velocities.
04

Calculate the center of mass velocity at 10 seconds

Using the formula for the velocity of the center of mass, insert the masses and velocities we already calculated: \(V_{CM} = \frac{(3.0\,\mathrm{kg})(98.1\,\text{m/s}) + (4.0\,\mathrm{kg})(98.1\,\text{m/s})}{3.0\,\mathrm{kg} + 4.0\,\mathrm{kg}}\) Calculate the result: \(V_{CM} = \frac{294.3\,\text{kg m/s} + 392.4\,\text{kg m/s}}{7.0\,\mathrm{kg}} = \frac{686.7\,\text{kg m/s}}{7.0\,\mathrm{kg}} = 98.1\,\text{m/s}\)
05

Report the CM velocity at 10 seconds

The velocity of the center of mass for the two falling bodies at \(t=10.0\,\mathrm{s}\) is \(98.1\,\text{m/s}\).

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Most popular questions from this chapter

Jane is sitting on a chair with her lower leg at a \(30.0^{\circ}\) angle with respect to the vertical, as shown. You need to develop a computer model of her leg to assist in some medical research. If you assume that her leg can be modeled as two uniform cylinders, one with mass \(M=\) \(20 \mathrm{kg}\) and length \(L=35 \mathrm{cm}\) and one with mass \(\quad m=\) \(10 \mathrm{kg}\) and length \(l=40 \mathrm{cm},\) where is the CM of her leg?
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