In a circus trapeze act, two acrobats actually fly through the air and grab on to one another, then together grab a swinging bar. One acrobat, with a mass of \(60 \mathrm{kg},\) is moving at \(3.0 \mathrm{m} / \mathrm{s}\) at an angle of \(10^{\circ}\) above the horizontal and the other, with a mass of $80 \mathrm{kg},\( is approaching her with a speed of \)2.0 \mathrm{m} / \mathrm{s}$ at an angle of \(20^{\circ}\) above the horizontal. What is the direction and speed of the acrobats right after they grab on to each other? Let the positive \(x\) -axis be in the horizontal direction and assume the first acrobat has positive velocity components in the positive \(x\)- and \(y\)-directions.

Step by step solution

01

Find the initial velocity components for each acrobat

First, we need to find the x and y components of the initial velocities of both acrobats. We can use trigonometry to do this: For the first acrobat (mass \(60 kg\)): \(v_{1x} = v_1 \cos{\theta_1} = 3.0 \mathrm{m/s} \cos{10^{\circ}}\) \(v_{1y} = v_1 \sin{\theta_1} = 3.0 \mathrm{m/s} \sin{10^{\circ}}\) For the second acrobat (mass \(80 kg\)): \(v_{2x} = v_2 \cos{\theta_2} = 2.0 \mathrm{m/s} \cos{20^{\circ}}\) \(v_{2y} = v_2 \sin{\theta_2} = 2.0 \mathrm{m/s} \sin{20^{\circ}}\)

02

Calculate the initial momenta

Next, we'll calculate the initial momenta of the acrobats in both the x and y directions: \(m_1 v_{1x} = 60 kg * v_{1x}\) \(m_1 v_{1y} = 60 kg * v_{1y}\) \(m_2 v_{2x} = 80 kg * v_{2x}\) \(m_2 v_{2y} = 80 kg * v_{2y}\)

03

Calculate the total momenta

Now we'll find the total momenta of the acrobats in the x and y directions: Total momentum in the x-direction: \(p_x = m_1 v_{1x} + m_2 v_{2x}\) Total momentum in the y-direction: \(p_y = m_1 v_{1y} + m_2 v_{2y}\)

04

Calculate the final velocities

Using the conservation of momentum, we can now calculate the final velocities in the x and y directions. Since they are grabbing on to each other, their combined mass is: \(m_t=m_1+m_2 = 60 kg + 80 kg = 140 kg\). Using the conservation of momentum: \(m_t v_{fx} = p_x\) and \(m_t v_{fy} = p_y\), so we can find their final velocities in the x and y directions: \(v_{fx} = \frac{p_x}{m_t}\) \(v_{fy} = \frac{p_y}{m_t}\)

05

Determine the direction and speed of the acrobats

Now, we can determine the direction, \(\alpha\), and the speed, \(v_f\), of the acrobats after they grab onto each other using basic trigonometry and the Pythagorean theorem: \(v_f = \sqrt{v_{fx}^2 + v_{fy}^2}\) \(\alpha = \tan^{-1}\left(\frac{v_{fy}}{v_{fx}}\right)\) Remember to plug in our numerical values of \(v_{fx}\) and \(v_{fy}\) into the equations above to obtain the final direction and speed of the combined acrobats.

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