F.4 Physics question

Method 1, by using the equation of motion:

Let a be the average ACCELERATION of the bullet inside the tree.

Initial velocity, u = 200 ms^-1
Final velocity, v = 0
Displacement, s = 5 cm = 0.05m

By the equation of motion,
v^2 = u^2 + 2as
(0)^2 = (200)^2 + 2a(0.05)
a = -400 000 ms^-2 (-ve acceleration implies deceleration)

Therefore, the average deceleration of the bullet inside the tree is 400 000 ms^-2 (deceleration should be +ve).


Method 2, by using energy:

By Newton's second law,
F = ma

By the conservation of energy,
Work done against the friction inside the tree = loss of K.E.
Fs = 1/2 mv^2 - 1/2 mu^2
ma(0.05) = 1/2 m(0)^2 - 1/2 m(200)^2
0.05a = 0 - 20 000
a = -400 000 ms^-2

Therefore, the average deceleration of the bullet inside the tree is 400 000 ms^-2 (deceleration should be +ve).

v=0, u=200ms^-1, s=0.05m
v^2-u^2=2as
-(200)^-2=2a(0.05)
a=-400000ms^-2
Therefore, the average deceleration of the bullet inside the tree is 400000ms^-2.