Disregarding air friction, what force(s) act on a bullet shot from a rifle as it moves through the air?
1 answer:
The gravity.
In fact, the gravity will "pull" the bullet towards the ground, and the magnitude of the force is equal to the weight of the bullet:
where
m is the mass of the bullet
is the acceleration of gravity
Due to the presence of this force, the bullet will have a parabolic motion, which consists of two independent motions on the horizontal axis and on the vertical axis:
- on the horizontal axis, the bullet moves by uniform motion with constant speed
- on the vertical axis, the bullet moves by uniformly accelerated motion, with constant acceleration g towards the ground.
In fact, the gravity will "pull" the bullet towards the ground, and the magnitude of the force is equal to the weight of the bullet:
where
m is the mass of the bullet
Due to the presence of this force, the bullet will have a parabolic motion, which consists of two independent motions on the horizontal axis and on the vertical axis:
- on the horizontal axis, the bullet moves by uniform motion with constant speed
- on the vertical axis, the bullet moves by uniformly accelerated motion, with constant acceleration g towards the ground.
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Answer:
a) K = 0.63 J, b) h = 0.153 m
Explanation:
a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is
w² =
where d is the distance from the pivot point to the center of mass and I is the moment of inertia.
The rod is a homogeneous body so its center of mass is at the geometric center of the rod.
d = L / 2
the moment of inertia of the rod is the moment of a rod supported at one end
I = ⅓ m L²
we substitute
w =
w =
w =
w = 4.427 rad / s
an oscillatory system is described by the expression
θ = θ₀ cos (wt + Φ)
the angular velocity is
w = dθ /dt
w = - θ₀ w sin (wt + Ф)
In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1
In the exercise it is indicated that at the lowest point the angular velocity is
w = 4.0 rad / s
the kinetic energy is
K = ½ I w²
K = ½ (⅓ m L²) w²
K = 1/6 m L² w²
K = 1/6 0.42 0.75² 4.0²
K = 0.63 J
b) for this part let's use conservation of energy
starting point. Lowest point
Em₀ = K = ½ I w²
final point. Highest point
Em_f = U = m g h
energy is conserved
Em₀ = Em_f
½ I w² = m g h
½ (⅓ m L²) w² = m g h
h = 1/6 L² w² / g
h = 1/6 0.75² 4.0² / 9.8
h = 0.153 m