Problem 96 - University Physics with Modern Physics

Question

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)?

Step-by-Step Solution

Answer

Piece 1 travels straight up at 3.00 m/s, and piece 2 travels straight back at 2.00 m/s.

Step 1: Determine the initial momentum of the bowling ball

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: \(p_{x0} = m \cdot v_{x0}\) \(p_{y0} = m \cdot v_{y0}\)

Step 2: Determine the final momentum of the pieces

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 \(p_{y3}\).

Step 3: Apply conservation of linear momentum

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: \(p_{x3} = p_{x0} - (p_{x1} + p_{x2})\) \(p_{y3} = p_{y0} - (p_{y1} + p_{y2})\)

Step 4: Find the velocity of the third piece

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.

Key Concepts

Momentum

Momentum is a fundamental concept in physics that describes the quantity of motion an object has. It's calculated as the product of an object's mass and velocity: \( p = m \cdot v \). In this exercise, the initial momentum of the bowling ball is determined using its mass (7.00 kg) and its velocity components derived from its launch speed and angle. Since momentum is a vector quantity, it has both magnitude and direction, which are broken down into horizontal and vertical components here. The conservation of momentum principle is crucial: it states that in a closed system, the total momentum remains constant unless acted upon by an external force. This principle helps us find the momentum and subsequently the velocity of the third piece post-explosion. It highlights how momentum transforms in isolated systems, providing insight into how objects will move after a force like an explosion is applied.

Velocity

Velocity is a vector quantity that includes both the speed and the direction of an object's motion. For the bowling ball, the initial velocity components are derived using trigonometric functions based on its speed and angle of projection. The velocity at launch is 10.0 m/s at an angle of 40.0 degrees. This velocity is broken into horizontal (\( v_{x0} = v_{0} \cos{40^{\circ}} \)) and vertical components (\( v_{y0} = v_{0} \sin{40^{\circ}} \)).

Understanding these components is critical as they allow us to apply the conservation of momentum to each direction separately. Post-explosion, to find the velocity of the third piece, the momentum in each direction is divided by the mass of one piece. This process effectively describes how the velocity changes during an explosion, giving insight into the origins of motion and direction for various fragments of an exploded object.

Explosion

An explosion is a rapid increase in volume and release of energy in an extreme manner, typically resulting in a physical change of the object involved. In this case, the explosion occurs at the peak of the bowling ball’s trajectory, fragmenting it into three equal parts. Each fragment follows its momentum based on how the forces of the explosion are distributed.

In this problem, because the explosion occurs at the peak, it doesn't affect vertical momentum initially, but it redistributes the momentum amongst the fragments. Two of the pieces' velocities are given: one moves upwards and one backwards (opposite to the original direction), and these known trajectories help in reshaping the conservation equations to solve for the unknowns. The explosion, being a sudden release, exemplifies how forces might act internally to change outputs post-event, yet obeying momentum conservation fully.

Trajectory

Trajectory is the path that an object follows through space as a function of time. When the bowling ball was launched, its initial velocity set a parabolic trajectory under the influence of gravity. The peak of this trajectory—the highest vertical point—is where the explosion happened, splitting the ball into three parts. Knowing part of the trajectory helps determine the subsequent paths of all fragments post-explosion.

Each piece follows a new trajectory based on its velocity after the explosion. The first travels upward, the second backward, and the third's path can be calculated using momentum conservation principles. The exercise reflects on the intrinsic characteristics of motion and trajectory, shedding light on concepts like path alteration due to internal forces, making it a vital practice to understand the dynamics of motion post-interaction or force application.