a bullet is fired into a trunk of a tree losses1/4 kinetic energy in travelling distance of 5cm .before stopping it travels a further distance of ....

Answer 1

ub^2 = .25 ua^2
average speed for initial slowing
= .5(ua+ub)

assume constant a
t = .05/.5(ua+ub) = .1/(ua+ub)

v = Vi +at
ub = ua + a t
ub-ua = a (.1)/(ua+ub)
.1 a = -(ua^2-ub^2)
.1 a = - (.75 ua^2)
a = -7.5ua^2

now complete trip
0 = ua + a t
t = ua/7.5ua^2 = 1/(7.5ua)
d = Vi t + .5 a t^2
d = ua/(7.5 ua)-.5 *7.5 ua^2 /(7.5ua)^2
d = .133 -.067 = .067 total distance
additonal distance = .067 - .050
= .017 m = 1.7 cm