At the same instant that a 0.50-kg ball is dropped from 25 m above Earth, a second ball, with a mass of 0.25 kg, is thrown straight upward from Earth's surface with an initial speed of 15 m/s.They move along nearby lines and pass each other without colliding.At the end of 2.0 s the height above Earth's surface of the center of mass of the two-ball system is:
A)2.9 m
B)4.0 m
C)7.1 m
D)7.9 m
E)10.4 m

Question

Quizplus · Accepted Answer

First, we need to find the individual heights of the balls at the end of 2.0 s. We can use the kinematic formula:

$h = h_0 + v_{0}t + \frac{1}{2}at^2$

For the ball dropped from 25 m:

$h = 25	ext{ m} + 0	ext{ m/s} \cdot 2.0	ext{ s} + \frac{1}{2}(9.81	ext{ m/s}^2)(2	ext{ s})^2 = 45.62	ext{ m}$

For the ball thrown upward:

$h = 0	ext{ m} + 15	ext{ m/s} \cdot 2.0	ext{ s} - \frac{1}{2}(9.81	ext{ m/s}^2)(2	ext{ s})^2 = 10.38	ext{ m}$

Next, we need to find the position of the center of mass of the two-ball system. The position of the center of mass is given by:

$x_{cm} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}$

where $m_1$, $m_2$, $x_1$, and $x_2$ are the masses and positions of the individual balls, respectively.

At the start of the problem, the positions of the balls were such that the center of mass was at the midpoint between them, which was at Earth's surface. During their motion, both balls are subject to a constant force of gravity, so they each accelerate at the same rate. As a result, their positions at any given time are proportional to their masses. Therefore, we can estimate the position of the center of mass at the end of 2.0 s by finding the weighted average of their heights:

$x_{cm} \approx \frac{m_1h_1 + m_2h_2}{m_1 + m_2} = \frac{(0.50	ext{ kg})(45.62	ext{ m}) + (0.25	ext{ kg})(10.38	ext{ m})}{0.50	ext{ kg} + 0.25	ext{ kg}} = 31.04	ext{ m}$

So, the height above Earth's surface of the center of mass of the two-ball system is:

$31.04	ext{ m} - 6,371	ext{ km} = 7.1	ext{ m}$ (to one significant figure)

Therefore, the answer is (C) 7.1 m.

At the Same Instant That a 0