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A 12.0-g rifle bullet is fired with a speed of 380 m/s into a ballistic pendulum with mass 6.00 kg, suspended from a cord 70.0 cm long (see Example 8.8 in Section 8.3). Compute (a) the vertical height through which the pendulum rises, (b) the initial kinetic energy of the bullet, and (c) the kinetic energy of the bullet and pendulum immediately after the bullet becomes embedded in the wood.

Short Answer

Expert verified
(a) 2.93 cm; (b) 867.6 J; (c) 1.72 J.

Step by step solution

01

Understand the Conservation Law Applicable

In this problem, we will use the law of conservation of momentum for the initial collision and conservation of energy after the bullet is embedded in the pendulum.
02

Apply Conservation of Momentum for the Collision

Conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. Initially, only the bullet has momentum. Using the formula: \( m_b \cdot v_b = (m_b + M) \cdot v_f \). Here, \( v_f \) is the final velocity of the system, \( m_b = 0.012 \) kg is the bullet's mass, \( v_b = 380 \) m/s is the bullet's velocity, and \( M = 6.00 \) kg is the pendulum's mass:\[ v_f = \frac{m_b \cdot v_b}{m_b + M} \]Substituting the values gives:\[ v_f = \frac{0.012 \times 380}{0.012 + 6} = 0.758 \text{ m/s} \]
03

Calculate the Vertical Height Using Conservation of Energy

After identifying \( v_f = 0.758 \) m/s, we use conservation of energy to find the height \( h \) the pendulum rises. The kinetic energy just after the bullet is embedded is converted to potential energy at the highest point:\[ \frac{1}{2} (m_b + M) v_f^2 = (m_b + M)gh \]Rearranging gives:\[ h = \frac{v_f^2}{2g} \]Using \( g = 9.8 \text{ m/s}^2 \):\[ h = \frac{(0.758)^2}{2 \times 9.8} = 0.0293 \text{ m} = 2.93 \text{ cm} \]
04

Compute the Initial Kinetic Energy of the Bullet

The initial kinetic energy (KE) of the bullet is found using the formula:\[ KE = \frac{1}{2} m_b v_b^2 \]Substituting the known values:\[ KE = \frac{1}{2} \times 0.012 \times (380)^2 = 867.6 \text{ J} \]
05

Find the Kinetic Energy after the Collision

The kinetic energy of the bullet-pendulum system right after the collision can be determined using:\[ KE = \frac{1}{2} (m_b + M) v_f^2 \]Substitute in the values:\[ KE = \frac{1}{2} \times (0.012 + 6) \times (0.758)^2 = 1.72 \text{ J} \]
06

Conclusion

The results are: (a) The vertical height through which the pendulum rises is 2.93 cm. (b) The initial kinetic energy of the bullet is 867.6 J. (c) The kinetic energy of the bullet and pendulum immediately after the collision is 1.72 J.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conservation of Momentum
In a ballistic pendulum scenario, conservation of momentum plays a crucial role during the collision phase. Momentum is the product of an object's mass and velocity, expressed as \( p = mv \). According to the principle of conservation of momentum, the total momentum of the system before a collision equals the total momentum after the collision. This principle helps us determine the velocity of the bullet-pendulum system immediately after the bullet becomes embedded in the pendulum.

When the bullet is fired into the pendulum, it carries momentum, while the pendulum is initially at rest and has no momentum. The equation for conservation of momentum can be set up as follows:
  • Before collision: Bullet momentum = \( m_b \cdot v_b \)
  • After collision: System momentum = \( (m_b + M) \cdot v_f \)
Where \( m_b \) is the bullet's mass, \( v_b \) is its velocity, \( M \) is the pendulum mass, and \( v_f \) is the final velocity after the collision. By solving \( m_b \cdot v_b = (m_b + M) \cdot v_f \), we can find \( v_f \), which is a stepping stone to further calculations regarding energy changes.
Conservation of Energy
Once the bullet is embedded in the pendulum, the conservation of energy principle comes into play. This principle states that energy cannot be created or destroyed but can only be transformed from one form to another. In this problem, the kinetic energy of the bullet-pendulum system is converted into potential energy as the pendulum rises to its highest point. This conversion allows us to calculate the height up to which the pendulum ascends.

Immediately after the collision, the system's energy is purely kinetic. As it swings upwards, this kinetic energy is transformed into gravitational potential energy. The conservation of energy equation can be laid out as:
  • Initial kinetic energy: \( \frac{1}{2} (m_b + M) v_f^2 \)
  • Potential energy at the highest point: \( (m_b + M)gh \)
By setting these energies equal \( \frac{1}{2} (m_b + M) v_f^2 = (m_b + M)gh \), and solving for \( h \), we can find the vertical height the pendulum reaches.
Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. It is defined mathematically by the expression \( KE = \frac{1}{2} mv^2 \). When considering a ballistic pendulum, kinetic energy explains the bullet's initial impact and the resulting motion of the pendulum. Understanding kinetic energy's role helps us quantify the energy involved in different stages of the event.

Initially, the bullet has kinetic energy given by \( KE = \frac{1}{2} m_b v_b^2 \). This energy allows the bullet to penetrate the pendulum. Once the bullet is embedded, the combined system's kinetic energy is calculated based on their final velocity \( v_f \). The drastic difference in kinetic energy before and after the collision reflects the energy loss mainly due to deformation and heat, demonstrating the inelastic nature of the collision.
Potential Energy
Potential energy in the context of a ballistic pendulum is gravitational potential energy. It is the energy held by an object because of its position relative to the Earth, calculated by \( PE = mgh \). After the collision, the ballistic pendulum swings upward, converting kinetic energy into potential energy at its peak height.

As the pendulum rises to its maximum height, its velocity momentarily drops to zero, and all the mechanical energy is stored as potential energy. This allows us to calculate how high the pendulum reaches by equating it with the kinetic energy just after the collision. Therefore, potential energy gives a clear measure of the system's energy state when motion halts during the swing, highlighting the energy transfer between types and the conservation throughout the system's motion.

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Most popular questions from this chapter

A 20.0-kg projectile is fired at an angle of 60.0\(^\circ\) above the horizontal with a speed of 80.0 m/s. At the highest point of its trajectory, the projectile explodes into two fragments with equal mass, one of which falls vertically with zero initial speed. Ignore air resistance. (a) How far from the point of firing does the other fragment strike if the terrain is level? (b) How much energy is released during the explosion?

Block \(B\) (mass 4.00 kg) is at rest at the edge of a smooth platform, 2.60 m above the floor. Block \(A\) (mass 2.00 kg) is sliding with a speed of 8.00 m/s along the platform toward block \(B\). \(A\) strikes \(B\) and rebounds with a speed of 2.00 m/s. The collision projects \(B\) horizontally off the platform. What is the speed of \(B\) just before it strikes the floor?

A small wooden block with mass 0.800 kg is suspended from the lower end of a light cord that is 1.60 m long. The block is initially at rest. A bullet with mass 12.0 \(g\) is fired at the block with a horizontal velocity \(v_0\). The bullet strikes the block and becomes embedded in it. After the collision the combined object swings on the end of the cord. When the block has risen a vertical height of 0.800 m, the tension in the cord is 4.80 N. What was the initial speed \(v_0\) of the bullet?

An 8.00-kg block of wood sits at the edge of a frictionless table, 2.20 m above the floor. A 0.500-kg blob of clay slides along the length of the table with a speed of 24.0 m/s, strikes the block of wood, and sticks to it. The combined object leaves the edge of the table and travels to the floor. What horizontal distance has the combined object traveled when it reaches the floor?

A 1200-kg SUV is moving along a straight highway at 12.0 m/s. Another car, with mass 1800 kg and speed 20.0 m/s, has its center of mass 40.0 m ahead of the center of mass of the SUV (\(\textbf{Fig. E8.54}\)). Find (a) the position of the center of mass of the system consisting of the two cars; (b) the magnitude of the system's total momentum, by using the given data; (c) the speed of the system's center of mass; (d) the system's total momentum, by using the speed of the center of mass. Compare your result with that of part (b).

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