91Ó°ÊÓ

Two African swallows fly toward one another, carrying coconuts. The first swallow is flying north horizontally with a speed of $20 \mathrm{m} / \mathrm{s} .$ The second swallow is flying at the same height as the first and in the opposite direction with a speed of \(15 \mathrm{m} / \mathrm{s}\). The mass of the first swallow is \(0.270 \mathrm{kg}\) and the mass of his coconut is \(0.80 \mathrm{kg} .\) The second swallow's mass is \(0.220 \mathrm{kg}\) and her coconut's mass is 0.70 kg. The swallows collide and lose their coconuts. Immediately after the collision, the \(0.80-\mathrm{kg}\) coconut travels \(10^{\circ}\) west of south with a speed of \(13 \mathrm{m} / \mathrm{s}\) and the 0.70 -kg coconut moves \(30^{\circ}\) east of north with a speed of $14 \mathrm{m} / \mathrm{s} .$ The two birds are tangled up with one another and stop flapping their wings as they travel off together. What is the velocity of the birds immediately after the collision?

Step by step solution

01

Find the total momentum before the collision

Let's first find the total momentum of each bird with its coconut before the collision. The momentum (p) of an object is given by the product of its mass (m) and velocity (v): \(p = mv\). For the first swallow: \(p_{1} = m_{1}v_{1} = (0.270+0.80)(20) = 1.07(20) = 21.4\,\mathrm{kg\,m/s}\). For the second swallow: \(p_{2} = m_{2}v_{2} = (0.220+0.70)(-15) = 0.92(-15) = -13.8\,\mathrm{kg\,m/s}\). The total momentum before the collision is given by \(p_{total} = p_{1} + p_{2} = 21.4 - 13.8 = 7.6\,\mathrm{kg\,m/s}\).

02

Find the total momentum after the collision

Let's now find the momentum of the coconuts after the collision. We need to break their momenta down to the x and y components, then add the components to determine the total momentum. For the \(0.80\,\mathrm{kg}\) coconut: \(p_{x1} = 0.80(13)\cos(10^{\circ}) = 10.40\,\mathrm{kg\,m/s}\) and \(p_{y1} = 0.80(13)\sin(10^{\circ}) = -2.25\,\mathrm{kg\,m/s}\). For the \(0.70\,\mathrm{kg}\) coconut: \(p_{x2} = 0.70(14)\sin(30^{\circ}) = 4.90\,\mathrm{kg\,m/s}\) and \(p_{y2} = 0.70(14)\cos(30^{\circ}) = 8.50\,\mathrm{kg\,m/s}\). The total momentum after the collision is given by \(p_{x} = p_{x1} + p_{x2} = 10.40 + 4.90 = 15.30\,\mathrm{kg\,m/s}\), and \(p_{y} = p_{y1} + p_{y2} = -2.25 + 8.50 = 6.25\,\mathrm{kg\,m/s}\).

03

Find the momentum of the birds after the collision

Using the conservation of momentum principle, the total momentum before the collision is equal to the total momentum after the collision. Now, let's find the momentum of birds after the collision by subtracting the momentum of the coconuts from the total momentum after collision in both the x and y directions. \(p_{bird_x} = p_{total} - p_{x} = 7.6 - 15.30 = -7.70\,\mathrm{kg\,m/s}\), and \(p_{bird_y} = p_{y} = 6.25\,\mathrm{kg\,m/s}\).

04

Find the velocity of the birds after the collision

Now that we have the momentum of the birds after the collision, we can find their velocity using their combined mass. The combined mass of the birds is given by \(m_{birds} = 0.270 + 0.220 = 0.490\,\mathrm{kg}\). The velocity components of the birds are given by \(v_{x} = \frac{p_{bird_x}}{m_{birds}} = \frac{-7.70}{0.490} = -15.7\,\mathrm{m/s}\), and \(v_{y} = \frac{p_{bird_y}}{m_{birds}} = \frac{6.25}{0.490} = 12.8\,\mathrm{m/s}\). The velocity of the birds immediately after the collision is given by the vector \((-15.7\,\mathrm{m/s}, 12.8\,\mathrm{m/s})\).

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