Register to reply 
Inelastic collision and kinetic energy lost. 
Share this thread: 
#1
Oct2309, 11:49 AM

P: 36

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? 


#2
Oct2309, 11:53 AM

P: 624

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
Oct2309, 12:00 PM

P: 36

How would I manipulate the variables to obtain this ratio?



#4
Oct2309, 12:04 PM

P: 624

Inelastic collision and kinetic energy lost.
Well, you have your COLM equation "(m1 + m2)Vf = m1Vi1 + m2Vi2" as you stated above, and correctly realised that "(m1 + m2)Vf = m1Vi1". Furthermore, you have the values for m1 and m2.



#5
Oct2309, 12:11 PM

P: 36

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
Oct2309, 12:18 PM

P: 624

Have you figured out how to relate this ratio to the % of initial kinetic energy carried off by the target(& projectile)? 


#7
Oct2309, 12:23 PM

P: 36

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
Oct2309, 12:32 PM

P: 624

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
Oct2309, 12:37 PM

P: 36

That is much easier... thank you so much!



#10
Nov2409, 05:41 PM

P: 57

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? 


Register to reply 
Related Discussions  
Inelastic collision, kinetic energy  Introductory Physics Homework  3  
Inelastic collision and Kinetic Energy Conservation  Introductory Physics Homework  2  
Kinetic energy of inelastic collision problem  Advanced Physics Homework  3  
Kinetic energy lost due to the collision  Introductory Physics Homework  0  
Collision (kinetic energy lost)  Introductory Physics Homework  5 