A 0.250kg bullet is initially at rest 2.00m above the earth in a rifle which is
ID: 2077638 • Letter: A
Question
A 0.250kg bullet is initially at rest 2.00m above the earth in a rifle which is aimed horizontally to the right. When the rifle fires the bullet, it forces the bullet with constant 500.0N force throughout rifle barrel which is 1.00m length to exit the rifle. a. Using the isolated system model, what is the total energy of the bullet while at rest in the rifle chamber b. Using the isolated system model, what is the kinetic energy of the bullet just as it exits the rifle barrel? c. What is the initial velocity of the bullet as it exits the rifle barrel? d. Assuming there is no air drag on the bullet, what is the final total kinetic energy when the bullet hits the ground?Explanation / Answer
a. FOrce, F = 500 N
Length, l = 1 m
then this is the energy stored as potential energy in the bullet, = 500*1 = 500 J
but we have ignored gravitational PE, = mgh = 0.25*9.81*2 = 4.905 J
Net PE = 504.905 J
b. From work energy theorem, this PE is converted to KE
KE = 500 J
c. KE = 0.5mv^2
500 = 0.5*0.25*v^2
v^2 = 4000
v = 63.245 m/s
d. KE at ground = 504.905 J ( from work eneergy theorem)
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