Bullet-Block Collision: Solving for Initial Speed and Energy Dissipation

In summary, a bullet of mass 1.4×10−3 {\rm kg} embeds itself in a wooden block with mass 0.999 {\rm kg}, compressing a spring (k = 110 {\rm N/m}) by a distance 5.5×10−2 {\rm m} before coming to rest. The initial speed of the bullet can be calculated using energy conservation, giving a value of 0.5767 m/s. The fraction of the bullet's initial kinetic energy that is dissipated in the collision can be determined using the equation 1/2mv^2=1/2mv^2-(work by friction+work by spring), but further steps are needed to solve for this
  • #1
bfusco
128
1

Homework Statement


A bullet of mass 1.4×10−3 {\rm kg} embeds itself in a wooden block with mass 0.999 {\rm kg}, which then compresses a spring (k = 110 {\rm N/m}) by a distance 5.5×10−2 {\rm m} before coming to rest. The coefficient of kinetic friction between the block and table is 0.46.
a)what is the initial speed of the bullet?
b)What fraction of the bullet's initial kinetic energy is dissipated (in damage to the wooden block, rising temperature, etc.) in the collision between the bullet and the block? (answer: ΔK/K
 
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  • #2
You must attempt to at a solution first before we can help you.
 
  • #3
tal444 said:
You must attempt to at a solution first before we can help you.

i did i didn't feel like typing it, but here you go.

i started by drawing a diagram labeling part A to be where the spring is compressed, part B to be where the block started pushing the spring, part C to be where the bullet impacts the block, and part D the firing of the bullet.
I then used energy conservation, stating that the E@A=E@B. 1/2kx^2=1/2mv^2, solving for v and getting sq.rt of (kx^2/m). plugging in the numbers i get v=.5767 m/s. Then i attempt to use energy conservation from B to C, 1/2mv^2=1/2mv^2-(work by friction+work by spring) but i kind of just put that equation together myself because i know energy is lost by work done) and i don't really know what to do.
 

FAQ: Bullet-Block Collision: Solving for Initial Speed and Energy Dissipation

1. What is conservation of momentum/energy?

The conservation of momentum/energy is a fundamental principle in physics that states that the total momentum/energy of a closed system remains constant over time, regardless of any internal changes or external forces acting on the system. This means that in any interaction or process, the total momentum/energy before and after the interaction remains the same.

2. How is conservation of momentum/energy related to Newton's laws of motion?

Newton's laws of motion are the basis for the conservation of momentum/energy. The first law states that an object will remain at rest or in motion at a constant velocity unless acted upon by an external force. This is related to the conservation of momentum/energy because when there is no external force acting on a system, its momentum/energy will remain constant.

3. What are some real-life examples of conservation of momentum/energy?

Some real-life examples of conservation of momentum/energy include a billiard game, where the total momentum/energy of the balls before and after a collision remains the same, and a rollercoaster, where the kinetic energy of the cart is converted into potential energy at the top of a hill and then back into kinetic energy as the cart moves down the hill.

4. Can conservation of momentum/energy be violated?

No, the conservation of momentum/energy is a fundamental law of physics and has been proven to hold true in countless experiments. There are no known cases where it has been violated.

5. How is conservation of momentum/energy used in engineering and technology?

The principles of conservation of momentum/energy are used extensively in engineering and technology, particularly in the design and operation of machines and systems. For example, in designing a car, engineers must ensure that the energy from the fuel is efficiently converted into the car's motion, and that the car's momentum is conserved in order to maintain control and stability while driving.

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