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 \(7.00-\mathrm{kg}\) 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)?

Short answer

The velocity of Fragment 3 is 9.87 m/s at an angle of 33.0°.

Step-by-step solution

Identify givens and unknowns

We are given: 1. The mass of the bowling ball (m) is 7 kg. 2. The initial speed (v) of the ball is 10 m/s. 3. The initial angle (θ) is 40°. 4. Fragment 1 has a velocity (v1) of 3 m/s upward. 5. Fragment 2 has a velocity (v2) of 2 m/s straight back. 6. we need to find the velocity (speed and direction) of Fragment 3.

Calculate the initial momentum

The initial momentum (p_initial) can be calculated using the mass of the ball (m) and its initial velocity (v). First, we need to find the initial velocity components (vx_initial and vy_initial) in the x and y directions. $$ vx_{initial} = v * \cos{θ} = 10 * \cos{40°} = 7.66 \mathrm{~m} / \mathrm{s} $$ $$ vy_{initial} = v * \sin{θ} = 10 * \sin{40°} = 6.43 \mathrm{~m} / \mathrm{s} $$ The momentum components (px_initial, py_initial) can be calculated by multiplying the initial velocity components (vx_initial, vy_initial) with the mass (m). $$ px_{initial} = m * vx_{initial} = 7 * 7.66 = 53.62 \, \mathrm{kg} \, \mathrm{m} / \mathrm{s} $$ $$ py_{initial} = m * vy_{initial} = 7 * 6.43 = 45.01 \, \mathrm{kg} \, \mathrm{m} / \mathrm{s} $$

Calculate the momenta of the first two fragments

The mass of each fragment will be one-third of the ball's mass (m/3). The momentum of Fragment 1 in the x and y directions: $$ px_{1} = 0 \, \mathrm{kg} \, \mathrm{m} / \mathrm{s} \\ py_{1} = (m/3) * v_{1} = (7/3) * 3 = 7.00 \, \mathrm{kg} \, \mathrm{m} / \mathrm{s} \\ $$ The momentum of Fragment 2 in the x and y directions: $$ px_{2} = -(m/3) * v_{2} = -(7/3) * 2 = -4.67 \, \mathrm{kg} \, \mathrm{m} / \mathrm{s} \\ py_{2} = 0 \, \mathrm{kg} \, \mathrm{m} / \mathrm{s} $$

Calculate the momentum of Fragment 3

The momentum of Fragment 3 in the x and y directions can be found using the conservation of momentum principle: $$ px_{3} = px_{initial} - px_{1} - px_{2} = 53.62 - 0 - (-4.67) = 58.29 \, \mathrm{kg} \, \mathrm{m} / \mathrm{s} $$ $$ py_{3} = py_{initial} - py_{1} - py_{2} = 45.01 - 7.00 - 0 = 38.01 \, \mathrm{kg} \, \mathrm{m} / \mathrm{s} $$

Calculate the velocity of Fragment 3

Now we can find the velocity components of Fragment 3 (vx_3, vy_3) by dividing its momentum components (px_3, py_3) by its mass (m/3): $$ vx_{3} = \frac{px_{3}}{m/3} = \frac{58.29}{7/3} = 8.34 \mathrm{~m} / \mathrm{s} $$ $$ vy_{3} = \frac{py_{3}}{m/3} = \frac{38.01}{7/3} = 5.43 \mathrm{~m} / \mathrm{s} $$ Finally, we can find the speed and direction of Fragment 3: $$ v_{3} = \sqrt{vx_{3}^{2} + vy_{3}^{2}} = \sqrt{(8.34)^2 + (5.43)^2} = 9.87 \, \mathrm{m} / \mathrm{s} $$ $$ \varphi = \arctan{\frac{vy_{3}}{vx_{3}}} = \arctan{\frac{5.43}{8.34}} = 33.0 ^{\circ} $$ Hence, the velocity of Fragment 3 is 9.87 m/s at an angle of 33.0°.

Key concepts

Momentum Conservation in Explosions

When dealing with the physics of explosions, one of the fundamental principles we rely on is the conservation of momentum. This principle states that momentum, a product of an object's mass and velocity, is conserved in a closed system, unless acted upon by an external force. In the context of an explosion, such as a firecracker shattering a bowling ball midair, this means the total momentum immediately before and after the event must be equal.

Momentum is a vector quantity, which means it has both magnitude and direction. During an explosion, the individual fragments may fly off in different directions with varying speeds, but when you mathematically combine (vectorially) all their momenta, you get a total momentum that matches the pre-explosion momentum. The key to solving such problems is breaking down the velocities into their component axes, typically 'x' for horizontal and 'y' for vertical, and applying the conservation law to each axis separately.

By understanding that the momentum of the system is the same before and after the explosion, students can set up equations that represent conservation on a component-by-component basis. This concept does not only apply to explosions but also to any collision or separation event where no external forces are interfering, such as in space.

Projectile Motion

Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory. The motion of a projectile is a two-dimensional motion, which can be decomposed into two independent motions: horizontal and vertical.

In our textbook exercise, the bowling ball is launched at an angle into the air, creating a parabolic trajectory. The horizontal motion is uniform, meaning the horizontal velocity remains constant if we neglect air resistance. In contrast, the vertical motion is uniformly accelerated, influenced only by the force of gravity, making the vertical velocity change over time.

Optimizing Projectile Motion Problems

Listed below are a series of tips that could help students tackle problems involving projectile motion:

• Break down initial velocity into horizontal and vertical components using trigonometry.
• Use kinematic equations to explore the vertical motion, keeping in mind that the acceleration due to gravity is a constant 9.81 m/s² downwards.
• Remember that at the peak of its motion, the vertical velocity of a projectile is zero.
• Horizontal distance covered can be calculated by multiplying the time of flight by the constant horizontal velocity.

Vector Components of Velocity

Velocity is another vector quantity in physics, requiring both a magnitude (speed) and a direction. In problems dealing with two dimensions, like the one with the exploding bowling ball, it's helpful to break down velocity into its horizontal (x) and vertical (y) components. This allows us to analyze the motion in each dimension separately, simplifying calculations and problem-solving.

The horizontal component of velocity remains constant when air resistance is negligible, while the vertical component of velocity changes due to the force of gravity. In the given problem, trigonometric functions like sine and cosine are used to find these components based on the launch angle and the initial speed.

Understanding Velocity Components

For better comprehension and problem-solving efficiency:

• Use cosine to find the horizontal component of velocity (adjacent to the angle).
• Use sine to find the vertical component of velocity (opposite to the angle).
• When combining velocity components to find the resultant velocity, use the Pythagorean theorem.
• The direction of the resultant velocity can be determined by the arctangent of the ratio between the vertical and horizontal components.

This methodology is crucial for decomposing velocity in multiple dimensions, which was necessary to find the velocity of the third fragment in our problem.