What is the velocity of an object that has a momentum of 4000 kg-m/a and a mass of 115 kg
2 answers:
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is the time interval elapses before the back end of the rod receives the message that it should stop.
Given:
Length of the rod, L = 80 cm = 0.800 m
Young's modulus, Y =
steel density,
The speed of the wave in the rod is,
Consequently, the length of the rod's end is traveled by the wave in at
Hence, is the time interval elapses before the back end of the rod receives the message that it should stop.
The three fundamental laws of classical mechanics known as Newton's laws of motion describe how an object's motion and the forces acting on it interact. The following paraphrase of these statutes is available
Unless a force acts upon a body, it remains at rest or in continual straight-line motion.
When a force acts on a body, the force is equal to the time rate at which the body's momentum changes.
When two bodies exert force on one another, the direction and amount of the force are opposed.
Isaac Newton first identified the three laws of motion in his 1687 book Philosophize Naturalis Principia Mathematica (Mathematical Principles of Natural Philosophy).
They served as the cornerstone for classical mechanics as Newton used them to examine and explain the motion of numerous physical objects and systems. The conceptual foundations of classical physics have been reconstructed in several ways since Newton, utilizing various mathematical techniques that have revealed insights that were hidden in the original, Newtonian formulation.
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Answer:
a) k = Mg / d
, b) v = √2gh
, c) v_{f} = , d) x² + 6d x -
dh = 0
e)the spring must compress a greater distance.
Explanation:
a) when the block of mass M is placed on the spring, we have an equilibrium condition,
∑ F = 0
- W = 0
k d = Mg
k = Mg / d
b) let's use the concepts of energy to find the velocity of the block just before the collision
starting point. Position when released
Em₀ = U = m g h
lowest point. Right at the point of shock
= K = ½ m v²2
as there is no friction, energy is conserved
Em₀ = Em_{f}
mg h = ½ m v²
v = √2gh
c) The velocity of the two blocks after the collision, we define a system formed by the two blocks, in such a way that the forces during the collision are internal and the moment is conserved
initial instant. Just before the crash
p₀ = 2M v + M 0
final instant. Just after the shock, before the spring compression begins
p_{f} = (2M + M) v_{f}
the moment is preserved
p₀ = p_{f}
2M v = 3M v_{f}
v_{f} = ⅔ v
v_{f} =
d) now we work with the joined system after the collision, let's use the concepts of energy
starting point. After shock, before beginning spring compression
Em₀ = K = ½ (3M)
Em₀ = 3/2 M (\frac{2}{3} \ \sqrt{2gh})²
Em₀ = 4/3 M gh
final point. With the spring fully compressed
Em_f = K_e + U = ½ k x² + (3M) g x
in this case we have taken the zero of gravitational potential energy at the point where the blocks collide, as there is no friction, the energy is conserved
Em₀ = Em_f
4/3 M g h = ½ k x² + 3M g x
½ k x² + 3Mg x - 4/3 Mgh = 0
we substitute the expression for k
(
) x² + 3Mg x -
Mgh = 0
+ 3 x -
h = 0
to find the value of the spring compression, the second degree equation must be solved
x² + 6d x - dh = 0
x = [-6d ± ] / 2
x = [-6d ± 6d ]/2
x = 3d ( -1± )
e) If the collision elastic force would not lose any part of the kinetic energy during the collision, therefore the speed of the block of mass M would be much higher and therefore the spring must compress a greater distance.