Inelastic collision and kinetic energy lost.

In summary, a projectile with a mass of 0.17 kg is fired at and embeds itself in a target with a mass of 2.43 kg. After being struck, the target (with the projectile in it) flies off. To find the percentage of the projectile's incident kinetic energy that the target carries off, we can use the equation (m1 + m2)Vf = m1Vi1 + m2Vi2 and manipulate the variables to obtain the ratio of Vf to Vi1. Alternatively, we can use the equation KE = (p^2)/2m to quickly calculate the ratio of kinetic energies. Both methods will result in a ratio of (m1)/(m1+m2), indicating
  • #1
IAmSparticus
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1. A projectile (mass = 0.17 kg) is fired at and embeds itself in a target (mass = 2.43 kg). The target (with the projectile in it) flies off after being struck. What percentage of the projectile's incident kinetic energy does the target (with the projectile in it) carry off after being struck?



2. (m1 + m2)Vf = m1Vi1 + m2Vi2
It is implied that the initial velocity of the target is 0 m/s, so the right side of the equation can be rewritten as m1Vi1+0.
Kinetic Energy= 1/2*mass*speed^2




3. Since the objects stick together, this is an inelastic collision. The total momentum before the collision and the total momentum after the collision are not going to be equal to each other. However, since we don't know the final velocity of the combined target and projectile or the initial velocity of the bullet, how are we supposed to figure this out?
 
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  • #2
IAmSparticus said:
3. Since the objects stick together, this is an inelastic collision. The total momentum before the collision and the total momentum after the collision are not going to be equal to each other. However, since we don't know the final velocity of the combined target and projectile or the initial velocity of the bullet, how are we supposed to figure this out?
No! There appears to be some confusion here; linear momentum is conserved for all collisions, elastic or inelastic (assuming no external forces). Inelastic collisions are simply collisions in which kinetic energy is not conserved.
You can manipulate the variables to obtain the ratio of vi to vf. Then try to use this ratio in the computation of the % of kinetic energy lost.
 
  • #3
How would I manipulate the variables to obtain this ratio?
 
  • #4
Well, you have your COLM equation "(m1 + m2)Vf = m1Vi1 + m2Vi2" as you stated above, and correctly realized that "(m1 + m2)Vf = m1Vi1". Furthermore, you have the values for m1 and m2.
 
  • #5
Ok so if I plug in the values of m1 and m2, I get an equation of 2.6kg Vf = .17 Vi1. Is this correct?

Now the next step I am a little confused on. If I divide both sides by 2.6 to find the ratio of Vf to Vi1, I get that Vf=.065Vi1. Is this correct or should I have done the opposite (find Vi1 to Vf)?
 
  • #6
IAmSparticus said:
Ok so if I plug in the values of m1 and m2, I get an equation of 2.6kg Vf = .17 Vi1. Is this correct?
Yup.
IAmSparticus said:
Now the next step I am a little confused on. If I divide both sides by 2.6 to find the ratio of Vf to Vi1, I get that Vf=.065Vi1. Is this correct or should I have done the opposite (find Vi1 to Vf)?
It doesn't matter which way, as long as you obtain a ratio between the two, you can solve the problem.
Have you figured out how to relate this ratio to the % of initial kinetic energy carried off by the target(& projectile)?
 
  • #7
Would it be that the final kinetic energy is 6.5% of the original kinetic energy, so 93.5% of the kinetic energy was lost? Or did I do it completely wrong?
 
  • #8
The numbers looks correct.
Anyway, here's an easier and faster way to do the problem using KE = (p^2)/2m (a useful identity to keep at hand.)
The ratio of the KEs would simply be (m1)/(m1+m2) since pf = pi.
 
  • #9
That is much easier... thank you so much!
 
  • #10
Momentum and Kinetic Energy are both conserved.
Assume a projectile velocity of 200m/s. Then the projectile will have 3400J of energy before impact. After impact the target and projectile have a mass of 2.6kg and a velocity of 13.08m/s. That equates to a total kinetic energy of 222.4J. If you take the original 3400J and subtract the KE required to slow the projectile down to 13.08m/s and subtract the KE required to speed up the target to 13.08m/s you will get 222.4J left over.
Doesn't that mean that KE and momentum are conserved?
 

1. What is an inelastic collision?

An inelastic collision is a type of collision between two objects where kinetic energy is not conserved. This means that after the collision, the total kinetic energy of the system is less than before the collision. In an inelastic collision, the objects stick together or deform upon impact, resulting in a loss of kinetic energy.

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

The amount of kinetic energy lost in an inelastic collision is affected by the masses and velocities of the objects involved. Generally, the larger the masses and the slower the velocities, the more kinetic energy will be lost in the collision.

3. How is the amount of kinetic energy lost in an inelastic collision calculated?

The amount of kinetic energy lost in an inelastic collision can be calculated using the following formula: Elost = 1/2(m1 + m2)(vf - vi)2Where m1 and m2 are the masses of the objects, vf is the final velocity after the collision, and vi is the initial velocity before the collision.

4. Can kinetic energy be gained in an inelastic collision?

No, kinetic energy cannot be gained in an inelastic collision. The total kinetic energy of the system must remain the same before and after the collision, but in an inelastic collision, some of the kinetic energy is converted into other forms of energy, such as heat or sound.

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

In an elastic collision, kinetic energy is conserved, meaning that the total kinetic energy of the system remains the same before and after the collision. The objects involved in an elastic collision bounce off each other without any loss of kinetic energy. In contrast, an inelastic collision does not conserve kinetic energy, resulting in a loss of kinetic energy as the objects stick together or deform upon impact.

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