The Role of the Spring in the Collision

In summary, the problem involves a 5.00-g bullet with an initial speed of 400 m/s passing through a 1.00-kg block connected to a spring of force constant 900 N/m. The block moves 5.00 cm to the right after impact. The first part of the problem involves finding the speed at which the bullet emerges from the block, which is 100 m/s. The second part involves finding the mechanical energy converted into internal energy in the collision, which is 374J. The spring's role in the problem is to help calculate the total change in kinetic energy, with the change in kinetic energy of the bullet being -374J and the change in kinetic energy of the spring being 1
  • #1
ch010308
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0

Homework Statement



A 5.00-g bullet moving with an initial speed of 400 m/s is fired into and passes through a 1.00-kg block, as in Figure P9.67. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring of force constant 900 N/m. If the block moves 5.00 cm to the right after impact, find
(a) the speed at which the bullet emerges from the block and
(b) the mechanical energy converted into internal energy in the collision.

Homework Equations



conservation of momentum
KE=1/2 mv2

The Attempt at a Solution



I solved part a, getting an answer of 100m/s for the speed at which the bullet emerges from the block. But I'm stuck at part b.
I interpreted the question as: finding the mechanical energy tat was converted = change in mechanical energy = change in kinetic energy of the bullet = 1/2 m(vf -vi) = (1/2)(5/1000)(100-400)2 = 225J

The correct answer should be 374J.

Pls help, thanks! :)
 

Attachments

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  • #2
What is the role played by the spring in this problem?
 
  • #3
hi

havenl't been through it all, but one thing that stands out is your last calc of cahnge in KE

should be
1/2 m(vf^2 -vi^2)

NOT
1/2 m(vf -vi)^@2as you've done in your calc
 
  • #4
should be
1/2 m(vf^2 -vi^2)

NOT
1/2 m(vf -vi)^@2as you've done in your calc

oh shoot! I can't believe I made tt mistake again n again. :tongue2: Thanks for pointing it out!

What is the role played by the spring in this problem?

ok, i think i get it.

change in KE of bullet = 1/2 m(vf2 -vi2)
= (1/2)(5/1000)(1002-4002) = -375J

change in KE of spring = (1/2)(1)(1.52 - 02) = 1.125J

total change in KE = -375 + 1.125 = -373.875 = -374J (round up to 3sf)

Is this correct? :)
 

1. What is linear momentum?

Linear momentum is a physical quantity that describes the motion of an object. It is defined as the product of an object's mass and velocity.

2. How is linear momentum conserved in a collision?

In a collision between two objects, the total linear momentum of the system remains constant. This means that the sum of the objects' individual momentums before the collision is equal to the sum of their momentums after the collision.

3. What is an elastic collision?

An elastic collision is a type of collision in which there is no loss of kinetic energy. This means that the total kinetic energy of the system before the collision is equal to the total kinetic energy after the collision.

4. Can linear momentum be negative?

Yes, linear momentum can be negative. This occurs when the direction of an object's velocity is opposite to the direction of its momentum. For example, if an object is moving to the left but its momentum is defined as positive to the right, then its momentum would be negative.

5. How is the coefficient of restitution related to linear momentum in a collision?

The coefficient of restitution is a measure of the elasticity of a collision. It is directly related to the change in linear momentum during a collision. A higher coefficient of restitution means a more elastic collision and a lower coefficient means a more inelastic collision.

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