/*! 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 41 A \(12.0-\mathrm{g}\) bullet is ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(12.0-\mathrm{g}\) bullet is fired horizontally into a \(100-\mathrm{g}\) wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant \(150 \mathrm{~N} / \mathrm{m}\). The bullet becomes embedded in the block. If the bullet-block system compresses the spring by a maximum of \(80.0 \mathrm{~cm}\), what was the speed of the bullet at impact with the block?

Short Answer

Expert verified
The final speed of the bullet at impact can be computed using the formula \(v_b=M*v_{ini}/m_b=sqrt{k*X^2/M}*sqrt{M/m_b}\) where all the values are given. Substitute all of them in to solve for \(v_b\).

Step by step solution

01

Determine the momentum before collision

The initial momentum of the bullet-block system is the mass of the bullet times the velocity of the bullet, as the block is initially at rest. Let's denote the mass of the bullet as \(m_b\), the velocity of the bullet as \(v_b\) and the mass of the block as \(m_{blk}\). The initial momentum \(P_{ini}\) is given by: \(P_{ini}=m_b*v_b\).
02

Determine the momentum after collision

After the bullet is embedded in the block, they become a single object moving together. Therefore, the momentum of the bullet-block system after the collision is the mass of the system times its velocity. It's worth noting that this is also the initial velocity with which the combined system compresses the spring. This velocity \(v_{ini}\) is what we need to find. The combined mass \(M = m_{blk} + m_b\) and the final momentum \(P_{final}=M*v_{ini}\). According to the law of conservation of momentum in a closed system, the momentum before the collision should equal the momentum after the collision, that is, \(P_{ini}=P_{final}\). From here, we can express \(v_{ini}\) as: \(v_{ini}=P_{ini}/M=m_b*v_b/M\).
03

Determine the speed of the bullet at impact

We now have everything we need to find the initial velocity of the bullet. Using the fact that the kinetic energy of the bullet-block system when the spring is at maximum compression is entirely converted from the initial kinetic energy, we can set up the equation \(1/2*M*v_{ini}^2=1/2*k*X^2\), where \(X\) is the maximum compression of the spring and \(k\) is the spring constant. From this, we can solve for \(v_b\), getting: \(v_b=M*v_{ini}/m_b=sqrt{k*X^2/M}*sqrt{M/m_b}\). Note that we have two square root terms here, but we know all of these values now and can simply substitute them in to solve for \(v_b\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. In our exercise, we see how kinetic energy plays a vital role in understanding the interactions within a bullet-block collision. The equation for kinetic energy is given by
  • \( KE = \frac{1}{2} m v^2 \)
where \( m \) is the mass and \( v \) is the velocity of the object. When the bullet is fired into the block, the system's total kinetic energy is initially related solely to the bullet, as the block is at rest. During the collision, the bullet's kinetic energy gets transferred and shared with the block. This energy transformation leads to compressing the spring. Post-collision, the kinetic energy of the bullet-block system converts to potential energy stored in the spring when it compresses to its maximum extent.
Spring Constant
The spring constant, denoted as \( k \), is an essential parameter in this problem. It measures a spring's stiffness. The higher the spring constant, the more force is needed to compress or stretch it by a given distance. In the given exercise, the spring constant is crucial for calculating the bullet's velocity post-collision. Hooke’s Law helps relate the force exerted on the spring to its displacement:
  • \( F = -kX \)
where \( F \) is the force exerted by the spring and \( X \) is the displacement. In our scenario, when the bullet-block system compresses the spring by \( 80 \) cm, the potential energy stored in the spring is
  • \( PE = \frac{1}{2}kX^2 \).
This potential energy is equal to the initial kinetic energy of the bullet-block system, allowing us to find
  • the speed of the bullet at impact.
Bullet-Block Collision
A bullet-block collision highlights the principles of the conservation of momentum and energy transformation. Before the impact, the bullet carries momentum while the block is at rest. Upon collision, the bullet embeds in the block, and they move together as a single unit. According to the
  • law of conservation of momentum: \( m_b \times v_b = M \times v_{ini} \).
This fundamental law asserts that without external forces, the system’s total momentum is conserved during the collision. The bullet's initial momentum turns into the system's momentum after they unite. Thus, this law allows us to calculate the post-collision velocity of the combined bullet-block system.
Velocity Calculation
Calculating the bullet's velocity at impact involves combining
  • the principles of conservation of momentum and energy.
Initially, the bullet had both momentum and kinetic energy. When the bullet embeds into the block, the resultant velocity post-collision can be calculated using momentum conservation, with the equation
  • \( v_{ini} = \frac{m_b \times v_b}{M} \).
Subsequently, during maximum spring compression, all kinetic energy is converted to potential energy in the spring, expressed as
  • \( \frac{1}{2}M \times v_{ini}^2 = \frac{1}{2}kX^2 \).
Solving these equations allows you to find
  • the bullet's velocity:
  • \( v_b = \sqrt{\frac{k \times X^2}{M}} \times \sqrt{\frac{M}{m_b}} \).
This methodical approach underscores how physics principles blend to unveil velocities in collision events.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A man of mass \(m_{1}=70.0 \mathrm{~kg}\) is skating at \(v_{1}=\) \(8.00 \mathrm{~m} / \mathrm{s}\) behind his wife of mass \(\mathrm{m}_{2}=50.0 \mathrm{~kg}\), who is skating at \(\tau_{2}=4.00 \mathrm{~m} / \mathrm{s}\). Instead of passing her, he inadvertently collides with her. He grabs her around the waist, and they maintain their balance. (a) Sketch the problem with before-and-after diagrams, representing the skaters as blocks. (b) Is the collision best described as elastic, inelastic, or perfectly inelastic? Why? (c) Write the general equation for conservation of momentum in terms of \(m_{1}, v_{1}, w_{2}, v_{2}\), and final velocity \(v\) - (d) Solve the momentum equation for \(v_{\gamma}\). (e) Substitute values, obtaining the numerical value for \(v_{f}\), their speed after the collision.

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 m, and the block slides a distance d before stopping. Find the initial speed v0 of the bullet in terms of M, m, m, g, and d.

S This is a symbolic version of Problem 33. A railroad car of mass \(M\) moving at a speed \(v_{1}\) collides and couples with two coupled railroad cars, each of the same mass \(M\) and moving in the same direction at a speed \(t_{2}\). (a) What is the speed \(v_{y}\) of the three coupled cars after the collision in terms of \(v_{1}\) and \(v_{2}\) ? (b) How much kinetic energy is lost in the collision? Answer in terms of \(M\), \(v_{1}\), and \(v_{2}\).

A billiand ball rolling across a table at \(1.50 \mathrm{~m} / \mathrm{s}\) makes a head-on elastic collision with an identical ball. Find the speed of each ball after the collision (a) when the second ball is initially at rest, (b) when the second ball is mowing toward the first at a speed of \(1.00 \mathrm{~m} / \mathrm{s}\), and (c) when the second ball is moving away from the first at a speed of \(1.00 \mathrm{~m} / \mathrm{s}\).

Calculate the magnitude of the linear momentum for the following cases: (a) a proton with mass equal to \(1.67 \times 10^{-27} \mathrm{~kg}\), moving with a speed of \(5.00 \times 10^{6}\) \(\mathrm{m} / \mathrm{s} ;\) (b) a \(15.0-\mathrm{g}\) bullet moving with a speed of \(300 \mathrm{~m} / \mathrm{s} ;\) (c) a \(75.0-\mathrm{kg}\) sprinter running with a speed of \(10.0 \mathrm{~m} / \mathrm{s}\); (d) the Earth (mass \(=5.98 \times 10^{24} \mathrm{~kg}\) ) moving with an orbital speed equal to \(2.98 \times 10^{4} \mathrm{~m} / \mathrm{s}\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.