After several large firecrackers have been inserted into its holes, a bowling ball is projected into the air using a homemade launcher and explodes in midair. During the launch, the 7.00 -kg ball is shot into the air with an initial speed of \(10.0 \mathrm{~m} / \mathrm{s}\) at a \(40.0^{\circ}\) angle; it explodes at the peak of its trajectory, breaking into three pieces of equal mass. One piece travels straight up with a speed of \(3.00 \mathrm{~m} / \mathrm{s}\). Another piece travels straight back with a speed of \(2.00 \mathrm{~m} / \mathrm{s}\). What is the velocity of the third piece (speed and direction)?

To solve this problem, we need to find the total initial momentum of the system consisting of the bowling ball. The initial momentum can be found using the formula: \(\text{momentum} = m \cdot v\) where \(m\) is the mass of the object, and \(v\) is the velocity. In this case, the mass of the bowling ball is 7.00 kg, and the initial speed is \(10.0 \mathrm{~m} / \mathrm{s}\) at a \(40.0^{\circ}\) angle above the horizontal. We can break this velocity into its horizontal and vertical components: \(v_{x0} = v_{0} \cos{40^{\circ}} = 10.0 \mathrm{~m} / \mathrm{s} \cdot \cos{40^{\circ}}\) (horizontal component) \(v_{y0} = v_{0} \sin{40^{\circ}} = 10.0 \mathrm{~m} / \mathrm{s} \cdot \sin{40^{\circ}}\) (vertical component) Now, we can find the initial momenta of the bowling ball in the horizontal (x) and vertical (y) directions: \(p_{x0} = m \cdot v_{x0}\) \(p_{y0} = m \cdot v_{y0}\)

After the explosion, the bowling ball splits into three equal-mass pieces. Let \(m_{piece} = \frac{7.00\,\text{kg}}{3}\). We are given the velocities of two of the pieces: Piece 1: travels straight up at \(3.00 \mathrm{~m} / \mathrm{s}\) Piece 2: travels straight back at \(2.00 \mathrm{~m} / \mathrm{s}\) To find their momenta, we can use the same formula as before. Remember that one piece is traveling vertically and the other is traveling horizontally in the opposite direction: \(p_{x1} = 0\) (piece 1 has no horizontal momentum) \(p_{y1} = m_{piece} \cdot 3.00 \mathrm{~m} / \mathrm{s}\) \(p_{x2} = - m_{piece} \cdot 2.00 \mathrm{~m} / \mathrm{s}\) (negative because it's traveling in the opposite horizontal direction) \(p_{y2} = 0\) (piece 2 has no vertical momentum) We're looking for the momentum of the third piece, \(p_{x3}\) and \(p_{y3}\).

Since no external forces act on the system, the total initial momentum equals the total final momentum. We can write this in the horizontal and vertical directions: \(p_{x0} = p_{x1} + p_{x2} + p_{x3}\) \(p_{y0} = p_{y1} + p_{y2} + p_{y3}\) We can now solve for the momentum of the third piece in the x and y directions: \(p_{x3} = p_{x0} - (p_{x1} + p_{x2})\) \(p_{y3} = p_{y0} - (p_{y1} + p_{y2})\)

We want the speed and direction of the third piece. To find the velocity components in the x and y directions, divide the found momenta by the mass of the third piece: \(v_{x3} = \frac{p_{x3}}{m_{piece}}\) \(v_{y3} = \frac{p_{y3}}{m_{piece}}\) Finally, we can find the speed and angle of the third piece using the magnitude and direction of the velocity vector: Speed: \(v_{3} = \sqrt{v_{x3}^{2} + v_{y3}^{2}}\) Direction: \(\theta_{3} = \tan^{-1}\left(\frac{v_{y3}}{v_{x3}}\right)\) (measured counterclockwise from the horizontal direction) This gives us the velocity (speed and direction) of the third piece after the explosion.