Inelastic Collision Calculate the Percent Change in KE

AI Thread Summary
An archer fires a 20g bolt at 105 m/s into a 400g apple at rest, with the bolt exiting at 45 m/s. The conservation of momentum equation is applied to find the final velocity of the apple, calculated to be 3 m/s. The change in kinetic energy (ΔKE) is determined to be -88,200 J, leading to a percent change in kinetic energy of -80%. A correction was noted regarding the isolation of variables in the momentum equation, clarifying the calculation steps. The solution was ultimately confirmed as correct.
arizona1379
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Homework Statement



An archer fires a 20g bolt with a speed of105 m/s through a 400 g apple initially at rest.
The bolt exits the apple with a speed of 45 m/s. What is the percent change in kinetic energy of the apple/bullet system?

m1= 20g
Vi1= 105 m/s
Vf2= 45 m/s
m2= 400g
Vi2= 0 m/s
Vf2= ?

Homework Equations



(m1v2+m2v2)i=(m1v2+m2v2)f

ΔKE= ((.5)(m1)(Vi1)2+(.5)(m2)(Vi2)2)i -((.5)(m1)(Vf1)2+(.5)(m2)(Vf2)2)f

%ΔKE= 100(F-I/I)

The Attempt at a Solution



(20g)(105 m/s)+(400g)(0 m/s)=(20g)(45 m/s)+ (400g)(vf)
2100(g)(m/s)=900(g)(m/s)+(400g)(vf)
1200g(g)(m/s)=400g(vf)
3 m/s = vf

ΔKE= ((.5)(20g)(105m/s)2+(.5)(400g)(0)2)i -((.5)(20g)(45m/s)2+(.5)(400g)(3 m/s)2)f

22050J-110250J=-88200J

100x(-88200)/110250J= -80%

Figured it out myself. :)

:::::Lashawnda:::::
 
Last edited:
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You did 2100 = 900 + 400 vf and isolated it wrong at the next step.
 
2100-900=1200
1200/400= 3
 
hjelmgart said:
you did 2100 = 900 + 400 vf and isolated it wrong at the next step.


2100-900=1200
1200/400= 3
 

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