Momentum Concervation and Collisions

In summary, the block slides a distance of 48.0 meters with the bullet's original velocity of 400ms^-1. After the collision, the bullet's velocity is reduced to 200ms^-1 and the block moves with a velocity of Vms^-1.
  • #1
mr.quakers
3
0

Homework Statement




A bullet with a mass of 7.00 , traveling horizontally with a speed of 400 , is fired into a wooden block with mass of 0.850 , initially at rest on a level surface. The bullet passes through the block and emerges with its speed reduced to 200 . The block slides a distance of 48.0 along the surface from its initial position.

What is the coefficient of kinetic friction between block and surface?

What is the kinetic energy of the block at the instant after the bullet passes through it?

Homework Equations



P=m1v1 + m2v2

KE = 1/2mv^2

...I think, I'm not completely sure on if it is elastic or inelastic. Our physics book is honestly terrible and has no examples relating to this.


The Attempt at a Solution



I really have no idea where to start. I've tried setting the momentums equal to each other and solving for v, but that doesn't work.

Second question:

A blue puck with a mass of 3.80×10−2 , sliding with a speed of 0.200 on a frictionless, horizontal air table, makes a perfectly elastic, head-on collision with a red puck with mass , initially at rest. After the collision, the velocity of the blue puck is 2.0×10−2 in the same direction as its initial velocity.

Find the magnitude of the velocity of the red puck after the collision.

Find the mass of the red puck.

Once again, no idea where to start.
 
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  • #2
mr.quakers said:

Homework Statement




A bullet with a mass of 7.00g , traveling horizontally with a speed of 400 ms^-1 , is fired into a wooden block with mass of 0.850kg , initially at rest on a level surface. The bullet passes through the block and emerges with its speed reduced to 200 . The block slides a distance of 48.0m along the surface from its initial position.

What is the coefficient of kinetic friction between block and surface?

What is the kinetic energy of the block at the instant after the bullet passes through it?
The letters in bold are what I assumed to be units.
Before collision:
Block has mass 0.850kg with a velocity of 0ms^-1
Bullet has mass 7g with a velocity of 400ms^-1

After collision:
Block moves with a velocity of Vms^-1
Bullet moves with velocity of 200ms^-1

Using the law of conservation of linear momentum
[tex]\frac{7}{1000}\times 400 + 0.850\times 0 = 0.850V+\frac{7}{1000}\times 200[/tex]

Is that what you did and it didn't work out?

Also an elastic collision is one in which kinetic energy is conserved.
 
  • #3
No that doesn't work. I have one more question, this section seems to be particularly bad at explaining the concepts.

A 22.00 lead sphere is hanging from a hook by a thin wire 3.80 long, and is free to swing in a complete circle. Suddenly it is struck horizontally by a 5.00 steel dart that embeds itself in the lead sphere.
 
  • #4
mr.quakers said:
No that doesn't work.

The velocity of the block is not obtained in that way? Well I don't see any other way to get it.


mr.quakers said:
A 22.00 lead sphere is hanging from a hook by a thin wire 3.80 long, and is free to swing in a complete circle. Suddenly it is struck horizontally by a 5.00 steel dart that embeds itself in the lead sphere

What exactly are you supposed to do in this one?
 
  • #5
I though I edited in the question, sorry. It's too late now though, the assignment is due in 15 minutes. Thanks for your help.
 

What is momentum conservation?

Momentum conservation is a fundamental principle in physics which states that the total momentum of a closed system remains constant, that is, it does not change over time. This means that in a collision or interaction between objects, the total momentum before the interaction is equal to the total momentum after the interaction.

How is momentum calculated?

Momentum is calculated by multiplying an object's mass by its velocity. The formula for momentum is p = mv, where p is momentum, m is mass, and v is velocity. Momentum is measured in units of kg m/s.

What are elastic and inelastic collisions?

An elastic collision is one in which both kinetic energy and momentum are conserved, meaning that the objects involved bounce off each other without any loss of energy. In contrast, an inelastic collision is one in which kinetic energy is not conserved, meaning that some energy is lost during the interaction between objects. In both cases, however, momentum is still conserved.

Why is momentum conservation important?

Momentum conservation is important because it is a fundamental law of nature that has been observed and tested in countless experiments. It allows scientists to predict the outcome of collisions and interactions between objects, and it also helps to explain the motion of objects in the universe.

How does momentum conservation apply to real-world situations?

Momentum conservation applies to many real-world situations, such as car crashes, billiard ball collisions, and even the motion of planets and stars. It is also an important concept in engineering, as it helps to design safer and more efficient structures and vehicles.

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