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Chapter 8: Problem 92

A 5.00 -g bullet is shot through a 1.00 -kg wood block suspended on a string 2.00 \(\mathrm{m}\) long. The center of mass of the block rises a distance of 0.38 \(\mathrm{cm} .\) Find the speed of the bullet as it emerges from the block if its initial speed is 450 \(\mathrm{m} / \mathrm{s}\) .

Short Answer

Expert verified
The bullet emerges with a speed of approximately 395.4 m/s.

Step by step solution

01

Identify the Known Values

The bullet has a mass \( m_b = 5.00 \) g or \( 0.005 \) kg. The wood block has a mass \( M = 1.00 \) kg. The length of the string is 2.00 m. The block's center of mass rises 0.38 cm or 0.0038 m. The initial speed of the bullet is \( v_{bi} = 450 \) m/s.
02

Applying Conservation of Momentum

Before the collision, only the bullet is moving with velocity \( v_{bi} \). After the collision, the bullet and block system moves together. By conservation of momentum, the initial momentum of the bullet equals the final momentum of the block and bullet system:\[m_b imes v_{bi} = (m_b + M) imes v_f \]where \( v_f \) is the final velocity of the system.
03

Calculate Final Velocity Using Rise Limit

The block rises a certain height due to kinetic energy converted to potential energy. By conservation of energy:\[\frac{1}{2} (m_b + M) v_f^2 = (m_b + M) g h \]Solve for \( v_f \):\[v_f = \sqrt{2gh}\]where \( g = 9.8 \) m/s虏 and \( h = 0.0038 \) m.
04

Solve for Final Velocity

Using \( h = 0.0038 \) m, calculate \( v_f \):\[v_f = \sqrt{2 imes 9.8 imes 0.0038} \\approx \sqrt{0.07444} \\approx 0.273 \text{ m/s}\]
05

Determine Speed of Bullet After Emergence

After determining \( v_f \), use the conservation of momentum equation to solve for the bullet's velocity after emerging, \( v_{bf} \):\[0.005 \times 450 = 0.005 \times v_{bf} + 1.00 \times 0.273 \]Isolating \( v_{bf} \):\[2.25 = 0.005 v_{bf} + 0.273 \1.977 = 0.005 v_{bf} \v_{bf} = \frac{1.977}{0.005} \v_{bf} \approx 395.4 \, \text{m/s}\]

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Key Concepts

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

Kinetic Energy
Kinetic energy is a measure of the energy an object has due to its motion. It is calculated using the formula \( K = \frac{1}{2} mv^2 \), where \( m \) is the mass and \( v \) is the velocity of the object. - In our exercise, the bullet, with its significant speed of 450 m/s, contains a substantial amount of kinetic energy before it hits the wood block.- After the bullet passes through the block, part of this kinetic energy is transferred to the block, causing it to move.- This transfer of energy is a critical part of understanding how objects interact and share energy during collisions.Understanding kinetic energy helps explain not just how fast objects move, but also how much work they are capable of doing or have already done during a collision. The higher the velocity and mass, the higher the kinetic energy.
Potential Energy
Potential energy is the energy stored by an object due to its position. In the context of our exercise, it is related to the height to which the block rises. - The block rises 0.38 cm because the kinetic energy transferred to it is converted into gravitational potential energy.- This potential energy can be calculated using \( PE = mgh \), where \( m \) is the mass of the block with the bullet inside, \( g \) is the acceleration due to gravity, and \( h \) is the height to which the block rises.In simple terms, when an object is lifted or elevated, it gains potential energy. This energy can be converted back into kinetic energy if the object is allowed to fall. For the block and bullet system, the potential energy is essential in understanding how much energy is conserved during the whole interaction.
Conservation of Energy
The principle of conservation of energy states that the total energy in an isolated system remains constant. Energy can change forms, such as from kinetic to potential, but it cannot be created or destroyed. In this exercise: - Initially, the bullet has kinetic energy as it speeds towards the block. - Upon passing through the block, some of this kinetic energy is transferred to the block, causing it to rise. - As the block rises, the kinetic energy changes to potential energy. This event demonstrates the conservation of energy as energy transfers and changes forms but remains constant in total. The principle helps us track energy transfers and transformations, reinforcing that while the speed and dynamics of individual components may change, the total energy within the system does not. This fundamental concept is crucial in solving such physics problems.

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Most popular questions from this chapter

An atomic nucleus suddenly bursts apart (fissions) into two pieces. Piece \(A,\) of mass \(m_{A},\) travels off to the left with speed \(v_{A} .\) Piece \(B,\) of mass \(m_{B},\) travels off to the right with speed \(v_{B}\). (a) Use conservation of momentum to solve for \(v_{B}\) in terms of \(m_{A}\) , \(m_{B},\) and \(v_{A}\) . (b) Use the results of part (a) to show that \(K_{A} / K_{B}=m_{B} / m_{A},\) where \(K_{A}\) and \(K_{B}\) are the kinetic energies of the two pieces.

On a very muddy football field, a 110 -kg linebacker tackles an \(85-\) kg halfback. Immediately before the collision, the line-backer is slipping with a velocity of 8.8 \(\mathrm{m} / \mathrm{s}\) north and the halfback is sliding with a velocity of 7.2 \(\mathrm{m} / \mathrm{s}\) east. What is the velocity (magnitude and direction) at which the two players move together immediately after the collision?

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

( a) Show that the kinetic energy \(K\) and the momentum magnitude \(p\) of a particle with mass \(m\) are related by \(K=p^{2} / 2 m\) (b) \(A 0.040\) -kg cardinal (Richmondena cardinalis) and a 0.145 -kg baseball have the same kinetic energy. Which has the greater magnitude of momentum? What is the ratio of the cardinal's magnitude of momentum to the baseball's? (c) A \(700-\mathrm{N}\) man and a \(450-\mathrm{N}\) woman have the same momentum. Who has the greater kinetic energy? What is the ratio of the man's kinetic energy to that of the woman?

Combining Conservation Laws. A 5.00 -kg chunk of ice is sliding at 12.0 \(\mathrm{m} / \mathrm{s}\) on the floor of an ice-covered valley when it collides with and sticks to another 5.00 -kg chunk of ice that is initially at rest. Fig. \(P 8.79\) ). Since the valley is icy, there is no friction. After the collision, how high above the valley floor will the combined chunks go?
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