Respuesta :
The definitions of momentum and impluse and their conservation allow finding the results for the different questions are:
1) The two quantities are in the same range
2) Elastic shock v =( 0.15 ±0.01 ) m/s
3) inelastic shock v = (0.30 ± 0.02) m / s
1) Momentum is a magnitude that indicates the change in the momentum of an object
I = ∫ F . dt = Δp
Where I is the momentum, F the force, t the time, Δp the change in momentum
Let's look for the average momentum
I = F Δt
I = m g Δt
I = 0.060 9.8 (2.30 - 1.1)
I = 0.71 kg m /s
The uncertainty is the statistical fluctuation of the magnitude, it is given by
ΔI = [tex]\frac{dI}{dt}[/tex]
ΔI = mg (Δt + Δt₀)
ΔI = 0.06 9.8 (0.02)
ΔI = 0.01
The result for momentum is
I = (0.71 ± 0.01) kg m / s
They ask to compare with the variation momentum
Δp = m v - m v₀
Let's calculate
Δp = 0.5 (1.5 -0)
Δp = 0.75 kg m / s
We calculate the uncertainty
Δp = [tex]\frac{dp}{dv}[/tex]
Δp = m (Δv + Δv)
Δp = 0.5 (0.2 + 0)
Δp = 0.1 kg m/s
The result for the variation momentum is
ΔP = (0.8 ± 0.2 ) kg m / s
We can see that the fluctuation of the momentum change is high, therefore it is within the fluctuation range of the impulse, the two quantities are in the same range
2) In this case we define a system formed by the two cars, for this system the momentum is conserved since the forces during the collision are internal
Initial instant. Before the crash
[tex]p_o = m_1 v_{1o}[/tex]
Final moment. After the crash
[tex]pf = m_1 v_{1f} + m_2v_{2f}[/tex]
The momentun is preserved
[tex]p_o = p_f\\m_1 v_{1o} = m_1 v_{1f} + m_2 v_{2f}[/tex]
As they indicate that the shock is elastic, the kinetic energy is conserved
Initial instant
K₀ = ½ m₁ [tex]v_{1o}^2[/tex]
Final moment
[tex]K_f[/tex] = ½ m₁ [tex]v_{1f}^2[/tex] + ½ m₂ [tex]v_{2f}^2[/tex]
Energy is conserved
[tex]K_o = K_f\\m_1 v_{1o}^2 = m_1 v_{1f}^2 + m_2 v_{2f}^2[/tex]
We write the system of equations
[tex]\frac{m_1}{m_2} (v_{1o} - v_{1f} ) = v_{2f}\\\frac{m_1}{m_2} (v_{1o}^2 - v_{1f}^2 ) = v_{2f}^2[/tex]
We solve the system of equations
( [tex]( v({1o}^2 - v_{1f}^2 ) = \frac{m_1}{m_2} \ ( v_{1o} - v(1f} )^2[/tex]
[tex]( v_{1o} + v_{1f} ) = \frac{m_1}{m_2} (v_{1o} - v_{1f} )\\v_{1f} ( 1 + \frac{m_1}{m_2} ) = v_{1o} ( \frac{m1}{m_2} - 1)[/tex]
We substitute
[tex]v_{1f} ( 1 + 2) =0.45 (2-1)\\v_{1f} = 0.15 m/s[/tex]
We look for uncertainty
Δv = [tex]\frac{d v_{1f}}{dv_{1o}}[/tex]
Δv = [tex]\frac{1}{3} \Delta v_{1o}[/tex]
Δv = 0.01 m/s
The result is
v =( 0.15 ±0.01 ) m/s
3) In this case the two vehicles remain united after the collision, inelastic shock
Let's use conservation of the moment
p₀ = p_f
m₁ v₁₀ = (m₁ + m₂) v
v = [tex]\frac{m_1}{m_1+m_2}[/tex] v₁₀
We calculate
v = [tex]\frac{0.6}{0.9 } \ 0.45[/tex]
v = 0.3 m / s
we look for uncertainty
Δv = [tex]\frac{d v}{d v_{1o}}[/tex]
Δv = [tex]\frac{m_1}{m_1+m_2} \ \Delta v_{1o}[/tex]
ΔV = [tex]\frac{0.6 }{0.9} \ 0.03[/tex]
Δv = 0.02 m / s
The result is
v = (0.30 ± 0.02) m / s
In conclusion with the definition of momentum and impulse and its conservation we can find the results for the different questions are:
1) The two quantities are in the same range
2) Elastic shock v =( 0.15 ±0.01 ) m/s
3) inelastic shock v = (0.30 ± 0.02) m / s
Learn more here: brainly.com/question/15652768
