A 0.053 kg bullet is fired into a 5.52 kg wood block, which is attached to a pendulum. At the instant after they collide inelastically, the velocity of the bullet + block system is 2.82 m/s. Assume that the height of the system at this time is h = 0 m.What is the total mechanical energy of the combined bullet and block system at this time?Hint: The following formula will be usefulTotal Energy = KE + PEWhere KE represents kinetic energy, and PE represents Potential Energy

Name: A 0.053 kg bullet is fired into a 5.52 kg wood block, which is attach
Uploaded: 2024-03-20T06:57:37.000Z
Channel: Bartleby
Description: A 0.053 kg bullet is fired into a 5.52 kg wood block, which is attached to a pendulum. At the instant after they collide inelastically, the velocity of the bullet + block system is 2.82 m/s. Assume that the height of the system at this time is h = 0

Answered: Image /qna-images/answer/79a07d06-6122-4bc8-9cfe-3cab90a98e4b.jpg

Science

Physics

A 0.053 kg bullet is fired into a 5.52 kg wood block, which is attached to a pendulum. At the instant after they collide inelastically, the velocity of the bullet + block system is 2 What is the total mechanical energy of the combined bullet and block system at this time? Hint: The following formula will be useful Total Energy = KE + PE Where KE represents kinetic energy, and PE represents Potential Energy

A 0.053 kg bullet is fired into a 5.52 kg wood block, which is attached to a pendulum. At the instant after they collide inelastically, the velocity of the bullet + block system is 2 What is the total mechanical energy of the combined bullet and block system at this time? Hint: The following formula will be useful Total Energy = KE + PE Where KE represents kinetic energy, and PE represents Potential Energy

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter8: Momentum And Collisions

Section: Chapter Questions

Problem 7P: Two blocks of masses m and 3m are placed on a frictionless, horizontal surface. A light spring is...

See similar textbooks

Similar questions

Consider an undamped linear oscillator with a natural frequency ω0 = 0.5 rad/s and the step function a = 1 m/s2. Calculate and sketch the response function for an impulse forcing function acting for a time τ = 2π/ω0. Give a physical interpretation of the results.
An inclined plane of angle = 20.0 has a spring of force constant k = 500 N/m fastened securely at the bottom so that the spring is parallel to the surface as shown in Figure P6.61. A block of mass m = 2.50 kg is placed on the plane at a distance d = 0.300 m from the spring. From this position, the block is projected downward toward the spring with speed v = 0.750 m/s. By what distance is the spring compressed when the block momentarily comes to rest?
Two blocks of masses m and 3m are placed on a frictionless, horizontal surface. A light spring is attached to the more massive block, and the blocks are pushed together with the spring between them (Fig. P8.7). A cord initially holding the blocks together is burned; after that happens, the block of mass 3m moves to the right with a speed of 2.00 m/s. (a) What is the velocity of the block of mass m? (b) Find the systems original elastic potential energy, taking m = 0.350 kg. (c) Is the original energy in the spring or in the cord? (d) Explain your answer to part (c). (e) Is the momentum of the system conserved in the bursting-apart process? Explain how that is possible considering (f) there are large forces acting and (g) there is no motion beforehand and plenty of motion afterward? Figure P8.7
In a laboratory experiment, 1 a block of mass M is placed on a frictionless table at the end of a relaxed spring of spring constant k. 2 The spring is compressed a distance x0 and 3 a small ball of mass m is launched into the block as shown in Figure P11.22. The ball and block stick together and are projected off the table of height h. Find an expression for the horizontal displacement of the ballblock system from the end of the table until it hits the floor in terms of the parameters given. FIGURE P11.22
A block is placed on top of a vertical spring, and the spring compresses. Figure P8.24 depicts a moment in time when the spring is compressed by an amount h. a. To calculate the change in the gravitational and elastic potential energies, what must be included in the system? b. Find an expression for the change in the systems potential energy in terms of the parameters shown in Figure P8.24. c. If m = 0.865 kg and k = 125 N/m, find the change in the systems potential energy when the blocks displacement is h = 0.0650 m, relative to its initial position. FIGURE P8.24
Check Your Understanding There is a second solution to the system of equations solved in this example (because the energy equation is quadratic): v1.f=-2.5m/s , v2.f=0 . This solution is unacceptable on physical grounds; what’s with it?
A 0.053 kg bullet is fired into a 5.38 kg wood block, which is attached to a pendulum. At the instant after they collide inelastically, the total mechanical energy of the system is 24.1 J. At the top of the swing of the pendulum, the block and bullet have risen to a maximum height, h, and now have a velocity of v = 0 m/s Use conservation of energy to find the maximum height, h, of the block + bullet system. Hint: The following formula will be useful PE = mx9.8×h %3D
A 10.0 g rifle bullet is fired with a speed of 380 m/s into a ballistic pendulum with mass 10.0 kg, suspended from a cord 70.0 cm long. Part A Compute the initial kinetic energy of the bullet. Express your answer in joules. Part B Compute the kinetic energy of the bullet and pendulum immediately after the bullet becomes embedded in the pendulum. Express your answer in joules. Part C Compute the vertical height through which the pendulum rises. Express your answer in centimeters.
You have to use the formula (mAvAi) + (mBvBi) = (mAvAf) + (mBvBf) With a stick that has a mass of 0.54kg, a quaranteen cracks a 4.4kg pinata when swinging the stick with a velocity of 4.8m/s. The stick stops after the collision and the pinata swings. What is the swing speed of the cracked pinata right after the collision?
. A 10-g bullet is fired into a ballistic pendulum with a mass of 6.00 kg and is hanging on a 0.70 m cord. When the bullet hits the pendulum it raises to a maximum height of 0.35 m. (a) Determine the initial speed of the bullet. Calculate the (a) kinetic energy of the bullet before the collision and the (c) kinetic energy of the bullet and pendulum system right after the bullet embeds itself into the pendulum.
An object with a weight of 5 N is left on the arc with a coefficient k = 2 N / cm from the height h. The spring is initially compressed by 3 cm through a plate connected by ropes from 2 ends. How much should the h be? Whether the collision will assume no loss of mechanical energy.
We can use conservation of momentum principles to compare the velocity of the marble before the collision to the velocity of the pendulum bob (with embedded marble) after the collision. After the collision, we can use conservation of energy principles to compare the initial kinetic energy of the (marble + pendulum) system with its potential energy when the pendulum reaches a maximum swing angle. Analyze the problem "backwards." Start with the motion of the (marble + pendulum) system after the inelastic collision. Using conservation of mechanical energy, derive an expression for the initial velocity of the marble + pendulum system as it starts to move. . Using trigonometry, you should be able to show that , where is the length of the ballistic pendulum, and is the maximum angle reached by the pendulum after the impact (show your work). Thus, the speed after the collision, , is