Bullet in a block hitting a spring

In summary, a bullet with a mass of 8 grams is shot into a 4.0 kg block on a frictionless surface. The bullet lodges in the block and the block then compresses a spring by 5.0 cm with a force constant of 2200 N/m. The initial velocity of the bullet is solved to be 0.686 m/s, but there was an error in the calculation as momentum and energy cannot be equal. The correct equation for kinetic energy is 1/2(m+M)v^2=1/2(k)(x)^2.
  • #1
preluderacer
84
0

Homework Statement



An 8-g bullet is shot into a 4.0-kg block, at rest on a frictionless horizontal surface. The bullet remains lodged in the block. The block moves into a spring and compresses it by 5.0 cm. The force constant of the spring is 2200 N/m. In the figure, the initial velocity of the bullet is closest to:


The Attempt at a Solution



what I did was set (mbullet+mblock)vblock=1/2(k)x^2 I solved the the velocity of the block and got 0.686 m/s. Then I set mbullet(v)bullet=(m1+m2)vblock and solved for the vbullet and got 343N. This answer doesn't seem quite right. What am I doing wrong?
 
Physics news on Phys.org
  • #2
preluderacer said:
what I did was set (mbullet+mblock)vblock=1/2(k)x^2 I solved the the velocity of the block and got 0.686 m/s.

The left-hand side of your equation is momentum, the right-hand side is energy. They can not be equal.

ehild
 
  • #3
i think i got it 1/2(m+M)v^2=1/2(k)(x)^2
 
Last edited:
  • #4
Write out the kinetic energy of the block with the bullet inside. ehild
 
  • #5


I would first check the units used in the calculation. It seems like there may be a mix-up of units, as the velocity of the bullet is given in Newtons (a unit of force) rather than meters per second (a unit of velocity). It is important to use consistent units in calculations to avoid errors.

Additionally, I would also check the equations and variables used in the calculation. It is important to use the correct equations and plug in the correct variables to get an accurate answer. In this case, the equation used should be conservation of momentum, rather than the equation for work and energy.

I would also double check the given values and make sure they are accurate and consistent. Sometimes, small errors in input values can lead to significant differences in the final answer.

Lastly, I would recommend checking the overall approach to the problem. Are all the necessary variables and factors being considered? Is there any assumption being made that may not be valid? It is always important to critically evaluate the approach to a problem to ensure accuracy.
 

1. What is the purpose of studying a bullet in a block hitting a spring?

The purpose of studying a bullet in a block hitting a spring is to understand the principles of conservation of momentum and energy. This concept is important in physics and has practical applications in fields such as engineering and ballistics.

2. How does the mass of the bullet and the block affect the motion of the spring?

The mass of the bullet and block will affect the motion of the spring by determining the amount of momentum and energy transferred to the spring upon impact. A heavier bullet or block will result in a larger transfer of momentum and energy, causing the spring to compress more.

3. How does the speed of the bullet affect the spring's compression?

The speed of the bullet will affect the spring's compression by determining the amount of kinetic energy transferred to the spring. A faster bullet will result in a larger transfer of energy, causing the spring to compress more.

4. Is the collision between the bullet and the spring perfectly elastic or inelastic?

The collision between the bullet and the spring is typically considered to be inelastic. This means that some of the kinetic energy is lost during the collision due to factors such as friction and deformation of the materials involved.

5. How does the stiffness of the spring affect the motion of the bullet and block?

The stiffness of the spring will affect the motion of the bullet and block by determining the amount of force required to compress the spring. A stiffer spring will require a larger force, resulting in a smaller compression distance. This can also affect the amount of energy transferred during the collision.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
3K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
4K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
9K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
3K
Back
Top