The conservation of energy and moment allows to find the results for the velocity of the pendulum and the bullet in the system are:
m (kg) v (m / s)
0.05 41.58
0.10 21.78
0.15 15.18
0.20 11.88
0.25 9.90
The conservation of mechanical energy is one of the most important principles of physics, it establishes that if there is no friction force, the energy is constant at all points. Mechanical energy is the sum of the kinetic energy plus the potential energies.
In the attachment we can see a diagram of the system, let's start by finding the speed of the pendulum when it leaves, for this we use the conservation of energy.
Starting point. In the lowest part of the movement.
Em₀ = K = ½ (m + M) v₀²
Final point. At the top of the movement.
Em_f = U = (m + M) g y
Energy is conserved because there is no friction.
Em₀ = Em_f
½ (m + M) v₀² = (m + M) g y
[tex]v_o = \sqrt{2gy}[/tex]
They indicate a table with several measurements of the masses and the period, let's use the relationship of the simple harmonic motion.
y = y₀ cos wt
The period and the angular velocity are related.
w = 2π / T
we substitute
y = y₀ cos ( [tex]2 \pi \frac{t}{T}[/tex] )
Let's analyze how long it takes to reach the point of maximum height, the period is the time of a complete oscillation, therefore from the lowest point to the highest point we have ¼ of oscillation, consequently the time to the highest point.
t = T / 4
y = y₀ cos ( [tex]2 \pi 4[/tex])
y = y₀
Therefore the point of maximum amplitude coincides with the maximum height and must be average by the student, suppose that the height is
y₀ = 20 cm = 0.20 m
Let's calculate the initial velocity.
v₀ = [tex]\sqrt{2 \ 9.8 \ 0.20 }[/tex]
v₀ = 1.98 m / s
They ask for the speed of the bullet before striking the basket of mass
M = 1 Kg.
The conservation of the momentum that for an isolated system the momentum is constant in all the instants. Let's form the system by the bullet and the basket.
Initial instant. Before the crash
p₀ = m v
Final moment. Right after the crash.
P_f = (m + M) v₀
The momentum is conserved because the system is isolated.
p₀ = p_f
m v = (m + M) v₀
v = m + M / m v₀
they have tabulated various mass for the bullet, we calculate the speed of each bullet.
m = 0.05 kg
v = [tex]\frac{0.05+1}{0.05} \ 1.98[/tex]
v = 41.58 m / s
m = 0.10 kg
v = [tex]\frac{0.10+1}{0.10} \ 1.98[/tex]
v = 21.78 m / s
m = 0.15 kg
v = [tex]\frac{0.15+1}{0.15y}[/tex]
v = 15.18 m / s
m = 0.20 kg
v = 11.88 m / s
m = 0.25 kg
v = 9.9 m / s
In conclusion with the conservation of energy and the momentum we can find the results for the speed of the pendulum and the bullet in the system are:
m (kg) v (m / s)
0.05 41.58
0.10 21.78
0.15 15.18
0.20 11.88
0.25 9.90
Learn more about conservation of energy and momentum here: brainly.com/question/25849204
