/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 75 ecp Measuring the speed of a bul... [FREE SOLUTION] | 91Ó°ÊÓ

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ecp Measuring the speed of a bullet. A bullet of mass \(m\) is fired horizontally into a wooden block of mass \(M\) lying on a table. The bullet remains in the block after the collision. The coefficient of friction between the block and table is \(\mu\), and the block slides a distance \(d\) before stopping. Find the initial speed \(v_{0}\) of the bullet in terms of \(M\), \(m, \mu, g\), and \(d\).

Short Answer

Expert verified
The initial speed of the bullet is \(v_{0} = \sqrt{\frac{2 \cdot m \cdot M \cdot g \cdot d}{\mu \cdot (M + m)^{2}}}\)

Step by step solution

01

Bullet-Block Collision

For the collision of the bullet and the block, momentum is conserved. If \(v_{0}\) is the initial velocity of the bullet and assuming the block is initially at rest, the total initial momentum is \(m \cdot v_{0}\). After the bullet gets lodged in the block, they move together with a common velocity \(v\). The final momentum is \((M + m) \cdot v\). From the conservation of momentum, \(m \cdot v_{0} = (M + m) \cdot v\). We can solve this equation for \(v = \frac{m \cdot v_{0}}{M + m}\).
02

Block-Bullet Sliding

Now consider the motion of the block (with the lodged bullet) sliding and coming to stop. The work done by friction is equal to the change in kinetic energy. The work done by friction is, \(W = -\mu \cdot M \cdot g \cdot d\). And the change in kinetic energy is, \(\Delta KE = \frac{1}{2} \cdot (M+m) \cdot v^{2} - 0\). Setting these two equal gives,-\(\mu \cdot M \cdot g \cdot d = \frac{1}{2} \cdot (M+m) \cdot v^{2}\). Solve this equation for \(v_{0} = \sqrt{\frac{-2 \cdot \mu \cdot M \cdot g \cdot d}{M+m}}\).
03

Substituting \(v\) for \(v_{0}\)

In this step, we substitute \(v = \frac{m \cdot v_{0}}{M + m}\) into \(v_{0} = \sqrt{\frac{-2 \cdot \mu \cdot M \cdot g \cdot d}{M+m}}\). Solving this equation for \(v_{0}\), you can find the initial velocity of the bullet, \(v_{0} = \sqrt{\frac{2 \cdot m \cdot M \cdot g \cdot d}{\mu \cdot (M + m)^{2}}}\).

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Key Concepts

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

Conservation of Momentum
In physics, the conservation of momentum is a crucial principle. It states that in a closed system, the total momentum is the same before and after an event. This means that no matter what happens within the system, the momentum remains unchanged—as long as no external forces act on it.
In our scenario where a bullet is fired into a block, the system consists of the bullet and the block. The bullet has an initial momentum, calculated as its mass (\(m\)) multiplied by its velocity (\(v_0\)).
When the bullet lodges into the block, the combined system of bullet and block move with a common velocity. The sum of their masses is \((M + m)\), with the final velocity being \(v = \frac{m \cdot v_{0}}{M + m}\).
  • The total momentum before collision: \(m \cdot v_0\)
  • The total momentum after collision: \((M + m) \cdot v\)
By equating these two values, the principle of conservation of momentum allows us to relate the initial speed of the bullet with the speed of the block and bullet after the collision.
Friction and Energy
Friction is the resistive force that opposes the motion of objects. It can convert kinetic energy into other forms, such as heat, preventing perpetual motion. This exercise involves friction because the block, with the embedded bullet, slides to a stop.
Sliding friction is quantified by the coefficient of friction \(\mu\) and is constant when the block moves. The frictional force can be calculated as \(\mu \cdot M \cdot g\), where \(M\) is the block's mass and \(g\) is the acceleration due to gravity. It steadily works against the block's kinetic energy, bringing it to rest after sliding a distance \(d\).
  • This frictional force does negative work on the block, \(-\mu \cdot M \cdot g \cdot d\).
  • This work against friction enables us to connect the kinetic energy immediately post-collision with the stopping distance of the block.
This connection between friction, energy conversion, and motion distance is vital for understanding how the energy in motion is dissipated.
Kinetic Energy Equation
Kinetic energy is the energy that an object possesses due to its motion and is given by the equation:\[KE = \frac{1}{2} m v^2\]
In our scenario, once the bullet embeds in the block, the whole system (block plus bullet) has kinetic energy determined by their combined mass and velocity:\[\text{KE} = \frac{1}{2} (M + m) v^2\]
During the bullet's embedding and block’s sliding, the initial kinetic energy in the system is transformed, primarily into thermal energy due to friction. The work done by friction equals this reduction in kinetic energy:
  • At the start, the kinetic energy is \(\frac{1}{2} (M + m) v^2\), representing the system’s motion post-collision.
  • As the block slides to a stop, the kinetic energy becomes zero.
By balancing the kinetic energy and the frictional work, the equation \(\frac{1}{2} (M + m) v^2 = \mu \cdot M \cdot g \cdot d\) helps solve for the initial speed \(v_0\) of the bullet.
Work-Energy Principle
The work-energy principle states that the net work done on a system is equal to the change in its kinetic energy. This principle is an essential concept in understanding how forces lead to changes in an object’s velocity.
In our exercise, after the bullet is lodged in the block, friction does work on the moving block-bullet system. Here’s a breakdown:
  • The work done by friction is negative because it acts opposite to the motion.
  • Initially, the moving block-bullet system has kinetic energy because of its velocity \(v\).
As friction performs work, it removes energy from the system, eventually bringing it to a halt with zero kinetic energy:\[\text{Work} = \Delta \text{KE} = \text{-}\mu \cdot M \cdot g \cdot d = \frac{1}{2} (M + m) v^2 - 0\]
Applying the work-energy principle, you can link the distance the block slides to the original bullet speed, demonstrating how energy initially in motion is dissipated through frictional work.

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Most popular questions from this chapter

A \(65.0\) -kg person throws a \(0.0450\) -kg snowball forward with a ground speed of \(30.0 \mathrm{~m} / \mathrm{s}\). A second person, with a mass of \(60.0 \mathrm{~kg}\), catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of \(2.50 \mathrm{~m} / \mathrm{s}\), and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged? Disregard friction between the skates and the ice.

A \(0.500-\mathrm{kg}\) football is thrown toward the east with a speed of \(15.0 \mathrm{~m} / \mathrm{s} .\) A stationary receiver catches the ball and brings it to rest in \(0.0200 \mathrm{~s} .\) (a) What is the impulse delivered to the ball as it's caught? (b) What is the average force exerted on the receiver?

A \(0.10\) -kg ball is thrown straight up into the air with an initial speed of \(15 \mathrm{~m} / \mathrm{s}\). Find the momentum of the ball (a) at its maximum height and (b) halfway to its maximum height.

An object has a kinetic energy of \(275 \mathrm{~J}\) and a momentum of magnitude \(25.0 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\). Find the speed and mass of the object.

An 8.00-g bullet is fired into a \(250-\mathrm{g}\) block that is initially at rest at the edge of a table of height \(1.00 \mathrm{~m}\) (Fig. \(\mathrm{P} 6.40)\). The bullet remains in the block, and after the impact the block lands \(2.00 \mathrm{~m}\) from the bottom of the table. Determine the initial speed of the bullet.

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