Conservation of Momentum problem needs quick reply

In summary: Plastic collision is a type of collision where the objects stick together instead of bouncing off each other. In summary, the problem involves a 24 g bullet being fired horizontally and embedding itself in a 10 kg block on a horizontal ice surface. The block slides and comes to rest after 2 seconds at a distance of 60 cm. The velocity of the bullet can be calculated using the conservation of momentum formula, but the answer may be affected by the frictional force, which can be found using Galileo's formula. However, this may not be necessary as the initial speed of the block can be calculated using the distance and time, assuming constant deceleration due to friction. There is also confusion about the use of Galileo's formula
  • #1
Byrne
20
0
Okay, here's the problem:

A 24 g bullet is fired horizontally, embedding itself in a 10 kg block initally at rest on a horizontal ice surface. The block slides the ice, coming to rest in 2.0 s at a distance of 60 cm from its original position. Assuming that the frictional force stopping the block was constant, calculate the velocity of the bullet.

Okay, I use the m1v1 + m2v2 = m1v1' + m2v2' formula to determine the v12'. However, my answer is half of what the answer in the back of the book is (My answer is 125 m/s while the book's is 250 m/s). Now, I think my problem might have something to do with the frictional force, but I don't really know where to begin to factor in the friction because the actual frictional force nor the frictional constant is given... please help.
 
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  • #2
U realize that this is a plastic collision & u can find the initial velocity of the system bullet through consevation of momentum...U can find the acceration due to friction using Galilei's formula.Then,knowing that this acceleration is constant,u can find the velocity of the system block-bullet and then the velocity of the bullet before impact (<--conservation of linear momentum).



Daniel.
 
  • #3
Well, only one out of three of the things you mentioned sound familiar to me. I mean, I am taking a high school physics course and we haven't learned anything about plastic collision or Galilei's formula so using those terms doesn't really help me.
 
  • #4
m1v1 + m2v2 = m1v1' + m2v2'

No, it's [tex]mv=(m+M)v'[/tex]

Since the bullet embeds itself inside the block.

Also, you don't need the constant of friction, since by knowing how far the block traveled and for how long you can figure out it's initial speed since the deaccelearation is constant (frictional force is constant).

EDIT:
Dextercioby was faster. :redface:
 
  • #5
Galileo's formula is valid for constant acceleration/deceleration movement and it reads

[tex] v_{final}^{2}=v_{initial}^{2}+2 a t [/tex]

For deceleration,a<0...

Plastic collision,u have to be kidding,right...?

Daniel.
 
  • #6
Okay, well the Galilei's forumula is obviously familiar, but as we learned it, its just one of the five acceration formulas. I guess my teacher doesn't care too much for the formal names of formulas. And I'm not kidding about plastic collision.

Thanks to both Berislav and dextercoiby for the help!
 
  • #7
I'm sorry guys, I've been trying to plug in these numbers for the last half an hour and cannot get the right answer.

You gave me Galileo's formula to use, but I still don't understand how to use it. I'm plugging in the info I do have and am not getting the answer... (200 m/s)
 
  • #8
[tex]v'=\frac{2s}{t}[/tex]
[tex]v'=0.6 m/s[/tex]

[tex]v=\frac{(m+M)v'}{m}[/tex]
[tex]v=250.6 m/s[/tex]
 

What is the conservation of momentum?

The conservation of momentum is a fundamental principle in physics that states that the total momentum of a closed system remains constant over time, regardless of any external forces acting on the system. This means that the total momentum before a collision or interaction will be equal to the total momentum after the collision or interaction.

How do you solve conservation of momentum problems?

To solve conservation of momentum problems, you will need to use the equation: m1v1 + m2v2 = m1v1' + m2v2', where m is the mass and v is the velocity of the objects before and after the collision. You will also need to make sure that the units for mass and velocity are consistent, and that the direction of the velocity is taken into account.

When is the conservation of momentum applicable?

The conservation of momentum is applicable to any closed system where there are no external forces acting on the system. This means that it can be applied to collisions between two objects, explosions, or any other interaction where the total momentum of the system remains constant.

What are some real-life examples of the conservation of momentum?

One example of the conservation of momentum is a billiard game. When the cue ball collides with another ball, the total momentum before the collision is equal to the total momentum after the collision. Another example is a rocket launching into space. The force of the rocket pushing against the ground creates an equal and opposite force that propels the rocket forward, conserving the total momentum of the system.

How does the conservation of momentum relate to Newton's third law of motion?

The conservation of momentum is closely related to Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. In a closed system, the forces acting on the objects will be equal and opposite, causing the total momentum of the system to remain constant. This is evident in the equation used to solve conservation of momentum problems, where the momentum of one object before the collision is equal to the momentum of the other object after the collision.

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