Respuesta :
Answer:
v₂ = -2.35 m/s ΔK = -0.364 J c)
Explanation:
a) For the first experiment, assuming no external forces acting during the collision, total momentum must be conserved.
⇒ Δp = 0 ⇒ p₀ = pf
p₀ = m₁*v₀₁ + m₂*v₀₂ = (.04 kg)* (2 m/s) + (.047 kg)* (-5 m/s) =-0.155 kg*m/s
pf = m₁*vf₁ + m₂*vf₂ = (0.04 kg)* (-1.11 m/s) + (0.047 kg)*(vf₂) = -0.155 kg*m/s
Solving for vf₂:
vf₂ = (-0.155 kg*m/s + 0.044 kg*m/s) / 0.047 kg = -2.35 m/s
b) We need to get first the initial kinetic energy of both masses, as follows:
K₀ = K₀₁ + K₀₂ = 1/2*(0.04 kg)(2m/s)² + 1/2*(0.047 Kg)*(-5m/s)² = 0.67 J
The final kinetic energy will be the sum of the final kinetic energies of both masses, as follows:
Kf = Kf₁ + Kf₂ = 1/2*(0.04 kg)(-3.8942 m/s)² + 1/2*(0.047 kg)*(0.0163 m/s)²
⇒ Kf = 0.303 J
⇒ ΔK = 0.303 J - 0.67 J = -0.364 J
In terms of loss of kinetic energy, the collision can be classified as inelastic, even though not completely, as both masses didn't stick together.
Answer:
a) v2 = -2.5319m/s
b) losses = 0.3642 J
Explanation:
Data Given:
[tex]m_{1} = 0.04 kg\\v_{1} = +2m/s , v'_{1} = -1.11m/s \\\\m_{2} = 0.047 kg\\v_{2} = -5m/s\\\\[/tex]
part a
Using conservation of momentum:
[tex]P_{f} = P_{i}\\m_{1}*v_{1} + m_{2}*v_{2} = m_{1}*v'_{1} + m_{2}*v'_{2} \\\\v'_{2} = \frac{m_{1}*v_{1} + m_{2}*v_{2} - m_{1}*v'_{1} }{m_{2}} \\\\v'_{2} = \frac{0.04*2 + 0.047*-5 - 0.04*-1.11 }{0.047} \\\\v'_{2} = -2.5319 m/s[/tex]
part b
Using energy conservation:
[tex]E_{1} + E_{2} = E'_{1} + E'_{2} + Losses\\\\Losses = E_{1} - E'_{1} + E_{2} - E'_{2}\\\\Losses = 0.5*m_{1}*(v^2_{1} - v' ^2_{1}) + 0.5*m_{2}*(v^2_{2} - v' ^2_{2}) \\\\Losses = 0.5*(0.04)*((2)^2 - (-3.8942)^2) + 0.5*(0.047)*((-5)^2 - (0.0163)^2)\\\\Losses = 0.3642 J[/tex]
