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Forensic science. Forensic scientists can measure the muzzle velocity of a gun by firing a bullet horizontally into a large hanging block that absorbs the bullet and swings upward. (See Figure 8.49.) The meas ured maximum angle of swing can be used to calculate the speed of the bullet. In one such test, a rifle fired a 4.20 g bullet into a 2.50 kg block hanging by a thin wire 75.0 \(\mathrm{cm}\) long. causing the block to swing upward to a maximum angle of \(34.7^{\circ}\) from the vertical. What was the original speed of this bullet?

Short Answer

Expert verified
The original speed of the bullet was approximately 947 m/s.

Step by step solution

01

Understand the Conservation of Momentum

First, note that when the bullet impacts the block, the system of bullet and block conserves momentum. Before the collision, only the bullet has momentum, while the block comes to rest together with the bullet right after the collision.
02

Express Conservation of Momentum

The conservation of momentum for the system can be expressed as:\[ m_{bullet} \times v_{bullet} = (m_{bullet} + m_{block}) \times v_{combined} \]where \( m_{bullet} \) is the mass of the bullet, \( v_{bullet} \) is the velocity of the bullet, \( m_{block} \) is the mass of the block, and \( v_{combined} \) is the velocity of the bullet-block system just after collision.
03

Convert Mass Units

Convert the bullet mass from grams to kilograms to keep units consistent:\[ m_{bullet} = 4.20 \text{ g} = 0.00420 \text{ kg} \]
04

Identify Energy Conservation during the Swing

When the block swings upward, the system's kinetic energy converts into gravitational potential energy at the maximum angle. Express this energy conversion as:\[ \frac{1}{2}(m_{bullet} + m_{block})v_{combined}^2 = (m_{bullet} + m_{block})gh \]
05

Calculate Maximum Height

The vertical height \( h \) can be calculated using the maximum angle \( \theta = 34.7^{\circ} \) and the length \( L \) of the wire:\[ h = L(1 - \cos \theta) \]Substitute the given values:\[ h = 0.75 \times (1 - \cos 34.7^{\circ}) \approx 0.75 \times (1 - 0.8290) \approx 0.128 \text{ m} \]
06

Substitute Energy Equation

Substitute the height into the potential energy equation:\[ \frac{1}{2}(m_{bullet} + m_{block})v_{combined}^2 = (m_{bullet} + m_{block})gh \]Solving for \( v_{combined} \), we have:\[ v_{combined} = \sqrt{2gh} \approx \sqrt{2 \times 9.8 \times 0.128} \approx 1.59 \text{ m/s} \]
07

Solve for the Initial Bullet Velocity

Using the conservation of momentum equation:\[ v_{bullet} = \frac{(m_{bullet} + m_{block})v_{combined}}{m_{bullet}} \]Substituting the known values:\[ v_{bullet} = \frac{(0.00420 + 2.50) \times 1.59}{0.00420} \approx 947 \text{ m/s} \]

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Key Concepts

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

Muzzle Velocity
Muzzle velocity is the speed of a bullet as it leaves the barrel of a gun. Knowing the muzzle velocity is crucial for understanding the projectile's behavior. In forensic science, this measurement can be performed by observing the bullet's effect on a pendulum setup. Here, the bullet is fired into a block, and the speed is deduced from the block's subsequent motion.

When the bullet hits the block, both objects move together. Despite the bullet losing speed and transferring its momentum to the block, the conservation of momentum allows us to calculate the original speed. This setup is not only valuable for forensic analysis but also illustrates classical principles of mechanics that deal with real-world situations.
Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. In our exercise, once the bullet becomes embedded in the block, their combined kinetic energy is a key piece of the overall analysis.

It is expressed as \( KE = \frac{1}{2}mv^2 \), indicating how energy scales with both mass and the square of velocity. Once the bullet is absorbed by the block, the total kinetic energy of the bullet-block system converts into potential energy as it swings upwards. This step highlights how energy transitions from one form to another — a core principle in physics. Understanding this process allows us to track and calculate kinetic energy changes through mechanical interactions such as collisions.
Gravitational Potential Energy
Gravitational potential energy relates to an object's position within a gravitational field. When the bullet-block system swings upward, its kinetic energy transforms into gravitational potential energy.

This transformation is represented by \( PE = mgh \), where \( m \) is mass, \( g \) is gravitational acceleration, and \( h \) is height. Calculating how high the block rises provides critical data to solve for the initial speed of the bullet. When the block achieves its maximum height, all kinetic energy is converted into potential energy, allowing us to backtrack through calculations and determine original speeds. This reflects the interconnectedness of energy types and their role in dynamics.
Energy Conservation
Energy conservation is a fundamental concept in physics stating that energy in an isolated system remains constant. During the bullet-block collision and the block's upward swing, the total mechanical energy transitions between kinetic and potential forms.

In this exercise, energy conservation helps us understand the process from collision impact to maximum height. The transformation follows strict laws; by ensuring calculations account for all energy forms involved, we maintain accurate solutions. By initially determining the system's combined kinetic energy and observing its potential energy at the swing's peak, we confirm that the conservation principle holds true. This principle not only demonstrates theoretical concepts but is practical in solving real-world physics problems.

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Most popular questions from this chapter

\(\bullet\) To warm up for a match, a tennis player hits the 57.0 g ball vertically with her racket. If the ball is stationary just before it is hit and goes 5.50 \(\mathrm{m}\) high, what impulse did she impart to it?

A bat strikes a 0.145 kg baseball. Just before impact, the ball is traveling horizontally to the right at \(50.0 \mathrm{m} / \mathrm{s},\) and it leaves the bat traveling to the left at an angle of \(30^{\circ}\) above horizontal with a speed of 65.0 \(\mathrm{m} / \mathrm{s}\) . (a) What are the horizontal and vertical components of the impulse the bat imparts to the ball? (b) If the ball and bat are in contact for 1.75 \(\mathrm{ms}\) , find the horizontal and vertical components of the average force on the ball.

Accident analysis. Two cars collide at an intersection. Car \(A\), with a mass of \(2000 \mathrm{~kg}\), is going from west to east, while car \(B\), of mass \(1500 \mathrm{~kg}\), is going from north to south at \(15 \mathrm{~m} / \mathrm{s}\). As a result of this collision, the two cars become enmeshed and move as one afterward. In your role as an expert witness, you inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of \(65^{\circ}\) south of east from the point of impact. (a) How fast were the enmeshed cars moving just after the collision? (b) How fast was car \(A\) going just before the collision?

\(\bullet\) Tennis, anyone? Tennis players sometimes leap into the air to return a volley. (a) If a 57 g tennis ball is traveling horizontally at 72 \(\mathrm{m} / \mathrm{s}\) (which does occur), and a 61 kg tennis player leaps vertically upward and hits the ball, causing it to travel at 45 \(\mathrm{m} / \mathrm{s}\) in the reverse direction, how fast will her center of mass be moving horizontally just after hitting the ball? (b) If, as is reasonable, her racket is in contact with the ball for \(30.0 \mathrm{ms},\) what force does her racket exert on the ball? What force does the ball exert on the racket?

\(\bullet\) A \(70-\mathrm{kg}\) astronaut floating in space in a \(110-\mathrm{kg}\) MMU (manned maneuvering unit) experiences an acceleration of 0.029 \(\mathrm{m} / \mathrm{s}^{2}\) when he fires one of the MMU's thrusters. (a) If the speed of the escaping \(\mathrm{N}_{2}\) gas relative to the astronaut is 490 \(\mathrm{m} / \mathrm{s}\) , how much gas is used by the thruster in 5.0 \(\mathrm{s} \%\) (b) What is the thrust of the thruster?

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