In a circus trapeze act, two acrobats fly through the air and grab onto each other, then together grab a swinging bar. One acrobat, with a mass of 54.0 kg, is moving at 3.00 m/s at an angle of 10.0° above the x-axis, and the other, with a mass of 84.0 kg, is approaching her with a speed of 2.00 m/s at an angle of 20.0° above the -x-axis. What is the speed of the acrobats right after they grab onto each other?

To find the speed of the acrobats right after they grab on to each other, we first need to determine their individual velocities in the x and y directions.

For the first acrobat (mass = 54.0 kg, initial velocity = 3.00 m/s, angle = 10.0° above the x-axis):

vx1 = 3.00 m/s

vy1 = 3.00 m/s * sin(10.0°) ≈ 0.530 m/s

For the second acrobat (mass = 84.0 kg, initial velocity = 2.00 m/s, angle = 20.0° above the -x axis):

vx2 = -2.00 m/s (since it's above the -x axis, the x-component is negative)

vy2 = 2.00 m/s * sin(20.0°) ≈ 1.149 m/s

Now, we can find their relative velocities in the x and y directions. Since they are grabbing each other, their x-component velocities will cancel out, and we only need to consider their y-component velocities:

vrelx = vx1 - vx2 = 3.00 m/s - (-2.00 m/s) = 5.00 m/s

vrelx = 0

vrey = vy1 - vy2 = 0.530 m/s - 1.149 m/s = -0.619 m/s

Now, we can find the total mass of both acrobats (m_total) and use their relative velocities to find their combined velocity after grabbing each other:

m_total = 54.0 kg + 84.0 kg = 138.0 kg

Using the conservation of momentum in the y-direction:

v_final_y * m_total = vrelx * (54.0 kg + 84.0 kg)

v_final_y = (vrelx * m_total) / (54.0 kg + 84.0 kg)

v_final_y = (-0.619 m/s * 138.0 kg) / 138.0 kg

v_final_y ≈ -0.619 m/s

Since the combined velocity in the y-direction is negative, we can conclude that the acrobats will slow down in the vertical direction after grabbing each other.

Now, we need to find the combined velocity in the x-direction after they grab each other. Since the x-component velocities cancel out, we can simply add the initial x-component velocities of both acrobats to find their combined velocity in the x-direction after grabbing each other:

v_final_x = vx1 + vx2

v_final_x = 3.00 m/s + (-2.00 m/s)

v_final_x = 1.00 m/s

Now, we can find the magnitude of the combined velocity (v_final) using the Pythagorean theorem:

v_final = √(v_final_x² + v_final_y²)

v_final = √((1.00 m/s)² + (-0.619 m/s)²)

v_final = √(1.00² + (-0.619)²)

v_final ≈ √(1.00 + 0.385201)

v_final ≈ √1.385201

v_final ≈ 1.174 m/s

So, the speed of the acrobats right after they grab on to each other is approximately 1.174 m/s.