Inelastic Collision Bullet/Block of Wood

In summary, the conversation is about a take home quiz where the student is working on a problem involving a bullet being fired into a block and causing it to swing out at an angle. The student has split the problem into two intervals and is considering using kinematics and angular motion to find the velocity of the bullet. However, after receiving input from others, the student realizes that conservation of mechanical energy can be used to solve the problem. The student still needs to find an unknown speed at the "join" of the two intervals.
  • #1
rmiller70015
110
1
Let me preface this by saying that this is a take home quiz and I'd really appreciate it if you didn't actually answer the question, I have outlined what I've done and what I plan to do to get an answer and I'm looking for input on my problem solving process.

I have a problem that I've been staring at for a while now and am not able to come up with a good starting place. The problem is:
A 20g bullet is fired into a 1000g block hanging from a 120cm long string. The bullet embeds itself into the block, and the block then swings out at an angle of 25 degrees. What is the speed of the bullet?

So far I've split it into two different time intervals, the first with the block at rest where Ft=-mg=0, and the second interval where the block is at 25 degrees with a velocity of zero. Then I resolved the forces acting on the block at each point. I'm just wondering where I should go from here. Using the inelastic collision equation I don't really get an answer because I don't know what v' is the instant after collision.

I guess the next step I would try is to use kinematics in 2 directions to find the instantaneous acceleration during impact, then use that to find the force exerted on the block by the bullet, and finally to find the velocity. Would this method get me where I'm going or is there something I'm missing?
 
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  • #2
hi rmiller70015! :smile:
rmiller70015 said:
Using the inelastic collision equation I don't really get an answer because I don't know what v' is the instant after collision.

call it v', then, and carry on :wink:

(and force is irrelevant)
 
  • #3
I don't know if I can call it v' since I'm looking for velocity to begin with, that leaves me with two variables. I'm also wondering if I can call v' the same velocity as what would be in the equation v=rω, but here I don't know my time interval.
 
  • #4
i'm not really following you :confused:

(and what does ω have to do with it anyway?)

you've divided the motion into two intervals

write out the collision equation for the first interval

what equation do you think you can use for the second interval?​
 
  • #5
I split it into two intervals for no real reason other than maybe I'd find something useful in there, but I don't think I did. As for what you're taking about the inelastic eqn is mAvA+mBvB=(mA+mB)v'. Since the only velocity I know is vB=0, I am left with v_A and v' as unknowns.

What I'm thinking at this point is to use angular motion, since the block would move along an arc.
 
  • #6
rmiller70015 said:
What I'm thinking at this point is to use angular motion, since the block would move along an arc.

since this is a take home quiz, i can't really tell you what equation to use for the second interval

(angular motion isn't going to help you at all)

what tells you how far up the pendulum will swing?​
 
  • #7
Hi rmiller70015,

Have you considered comparing the energy before and after the collision?
 
  • #8
After perusing through the course material I found that while I can't actually use conservation of energy on an elastic collision, I can use the conservation of mechanical energy for the time after the projectile stops deforming the block of wood, until the point where the wood/bullet system is at v=0, thank you for all your help here.
 
  • #9
hi rmiller70015! :smile:
rmiller70015 said:
After perusing through the course material I found that while I can't actually use conservation of energy on an elastic collision, I can use the conservation of mechanical energy for the time after the projectile stops deforming the block of wood, until the point where the wood/bullet system is at v=0

(you mean inelastic :wink:)

that's correct …

you can always split the motion into two parts, one "during" the collision (in which mechanical energy is not conserved), and one "after" (in which it is) :smile:

(and of course you have to join them together by using an unknown speed v at the "join")
 

1. What is an inelastic collision?

An inelastic collision is a type of collision in which the total kinetic energy of the system is not conserved. This means that some of the initial kinetic energy is lost during the collision, usually due to the deformation of the objects involved.

2. How does an inelastic collision differ from an elastic collision?

In an elastic collision, the total kinetic energy of the system is conserved. This means that the objects involved bounce off each other without any loss of energy. In an inelastic collision, some of the initial kinetic energy is converted into other forms of energy, such as heat or sound.

3. How does a bullet/block of wood inelastic collision occur?

A bullet/block of wood inelastic collision occurs when a bullet hits a block of wood and becomes embedded in it, instead of bouncing off. This results in a loss of kinetic energy and deformation of the bullet and block of wood.

4. What factors affect the amount of energy lost in an inelastic collision?

The amount of energy lost in an inelastic collision depends on factors such as the materials and shapes of the objects involved, the speed and angle of impact, and the presence of any external forces. Objects that are more deformable tend to result in a greater loss of energy.

5. How is the loss of energy in an inelastic collision calculated?

The loss of energy in an inelastic collision can be calculated by finding the difference between the initial kinetic energy and the final kinetic energy of the system. This can be determined using the formula: ΔK = Ki - Kf, where ΔK is the change in kinetic energy, Ki is the initial kinetic energy, and Kf is the final kinetic energy.

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