Solving for Initial Bullet Speed: Mass, Spring, and Friction

In summary: The velocity is = to .9495 m/s, I found this using .5mv^2 - force of friction*distance the spring compressed=.5kx^2.
  • #1
zyphriss2
18
0
A bullet of mass 1.6×10−3 {\rm kg} embeds itself in a wooden block with mass 0.990 {\rm kg}, which then compresses a spring (k = 160 {\rm N/m}) by a distance 5.0×10−2 {\rm m} before coming to rest. The coefficient of kinetic friction between the block and table is 0.51. What is the initial speed of the bullet? I have been working on this for and hour and a half and have done all of the other 8 problems except this one. Can anyone please help me and explain how it works.
 
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  • #2


Show what you've done and where you got stuck.

Hint: Work backwards. How fast was the bullet+block moving just after the collision?
 
  • #3


I have tried working backwards. I found that for the spring to be compressed the stated length that the block+bullet must exert .2 Newtons of force on the spring. Then i found that the summ of the forces in the x direcction is (block+bullet)a-force of friction=.2 and that results in a acceleration of 5.19. Now from here i do not know where to go
 
  • #4


For the spring compression part of this problem, use energy conservation, not force analysis.
 
  • #5


I had initially tried that. .5kx^2=.5mv^2 which turned out to make velocity equal to .635 m/s. Now from there I didn't know where to go, the coefficient of friction is messing me up.
 
  • #6


zyphriss2 said:
I had initially tried that. .5kx^2=.5mv^2 which turned out to make velocity equal to .635 m/s. Now from there I didn't know where to go, the coefficient of friction is messing me up.
You set KE = Spring PE, which isn't true since some of that original energy was lost to friction. You need to take the initial KE and subtract the work done by friction to get the final spring PE.
 
  • #7


I found that the velocity is = to .9495 m/s, I found this using .5mv^2 - force of friction*distance the spring compressed=.5kx^2. is this correct and from here would I just plug this velocity into mbullet*velocitybullet= m(bullet+block)*.9495?
 
  • #8


Thank you very much I got it correct.
 

1. How does mass affect the initial bullet speed?

Mass affects the initial bullet speed by influencing the amount of force needed to accelerate the bullet. The larger the mass, the more force is required to accelerate the bullet to a certain speed. This is due to the principle of inertia, which states that objects with more mass require more force to change their state of motion.

2. What role does the spring play in determining the initial bullet speed?

The spring plays a crucial role in determining the initial bullet speed. It is responsible for providing the force needed to accelerate the bullet. The more compressed the spring is, the greater the force it can exert on the bullet, resulting in a higher initial speed.

3. How does friction affect the calculation of initial bullet speed?

Friction works against the force provided by the spring, slowing down the acceleration of the bullet. This means that the higher the friction, the lower the initial bullet speed will be. It is important to take into account the level of friction in order to accurately calculate the initial bullet speed.

4. Can the initial bullet speed be calculated using only mass and spring constant?

No, the initial bullet speed cannot be calculated using only mass and spring constant. The level of friction also needs to be taken into account in order to accurately calculate the initial bullet speed. Additionally, the design and mechanics of the gun itself can also impact the initial bullet speed.

5. Is there a standard equation for solving for initial bullet speed?

Yes, there is a standard equation for solving for initial bullet speed. It is known as the spring-mass equation and is expressed as V = √(k/m)*(x₀-x), where V is the initial bullet velocity, k is the spring constant, m is the mass of the bullet, x₀ is the initial compression of the spring, and x is the final compression of the spring.

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