Im sure Imma mess this up, probably because Im really tired....but I have a bullet that weights .015kg's and is moving at 285m/s, so 609.18J's as with KE. I hit a 70kg block, penetrates .3048 meters thus exerting 1998.62N's of force, which is a lot of force. .3048/285m/s so 0.00106 seconds, now we can that this with impulse of force for... Favg•∆t = m • ∆v -> (1998.62N's)•(0.00106) = 2.1374N•s impulse of force, now we use this to calculate the change in velocity ∆v= 2.1374N•s/ .015kg = ∆v:142.5m/s Vi:285m/s Vf:427.5 m/s .015kg • 142.5 = 2.1374 N•s If I take my impulse of 2.1374N•s over 70kg's...the velocity is .03053m/s and we divide by 0.00106 seconds we get 28.55 m/s^2 acceleration.... We can rework this obviously, 1998.62N's over 70kg's = 28.55m/s d=(0m/s)(.00106s)+1/2(28.55m/s^2)(0.00106s^2) d=0.000016328 meters Object of 70kg Vf=(0m/s)(s)+(28.55m/s)(0.00106s) Object of 70kg Vf=0.03053m/s momentum p=70kg • 0.03053m/s = 2.13kg-m/s KE=0.032622 Joules..... The force applied by bullet is constant.... So, what type of collision is this? why did the Vf of the bullet become 427 when it is stopped....and Im doing this correct RIGHT? This is a concept check, not homework.