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 ....
To find the further distance the bullet travels before stopping, we can use the principle of conservation of energy.
Given that the bullet loses 1/4 of its kinetic energy in traveling a distance of 5 cm, we can say that 3/4 of its initial kinetic energy is still remaining.
Let's assume the further distance the bullet travels before stopping is 'x'.
Since kinetic energy is directly proportional to the square of the velocity, and the bullet loses 1/4 of its kinetic energy, the remaining kinetic energy can be represented as (3/4)^2 = 9/16.
So, the kinetic energy at the further distance is (9/16) of the initial kinetic energy.
Now, if we assume the initial kinetic energy to be K, then the kinetic energy at the further distance is (9/16)K.
Since energy is conserved, the remaining kinetic energy (9/16)K is equal to the kinetic energy at the further distance.
Therefore, we can say that (9/16)K = (1/2)m(v^2), where m is the mass of the bullet and v is the final velocity.
Now, let's solve for v:
(9/16)K = (1/2)m(v^2)
(9/16) = (1/2)(v^2)
Multiplying both sides by 16/9:
1 = 8/9(v^2)
(v^2) = 9/8
Taking the square root of both sides:
v = √(9/8)
v = (3/2)√2
Now, we know that the final velocity (v) is given by the formula v = u + 2as, where u is the initial velocity, a is the acceleration, and s is the distance traveled. In this case, the bullet starts with an initial velocity of u = 0 and comes to a stop, so the final velocity (v) will also be 0.
0 = 0 + 2a(x + 0.05)
0 = 2ax + 0.1a
2ax = -0.1a
2x = -0.1
x = -0.1/2
x = -0.05
The negative value suggests that the bullet does not travel further; it comes to a stop after traveling a distance of 5 cm.
Therefore, the further distance the bullet travels before stopping is 5 cm.
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