Respuesta :
Answer:
Explanation:
For elastic collision , the formula for velocity of .06 kg is
v₁ = (m₁-m₂)/(m₁+m₂) u₁ + 2m₁m₂/(m₁+m₂) u₂
=( .06-.09 / .06+.09 ) 5.5 + (2 x .06 x .09 / .06+.09 ) 3.4
=( -.03 / .15) x 5.5 + (2 x .0054 / .15) x 3.4
= -1.1 +.2448
= - 0 .8552 m/s
Its direction will be - opposite direction
the formula for velocity of .09kg is
v₂ = (m₂-m₁)/(m₁+m₂) u₂ + 2m₁m₂/(m₁+m₂) u₁
= ( .09-.06 / .06+.09 ) 3.4 + (2 x .06 x .09 / .06+.09 ) 5.5
= .68 +.396
= 1.076 m / s
Its direction will be in the same direction .
The conservation of the momentum and the kinetic energy allows to find the results for the elastic collision of the two masses is:
A. The speed of the lighter mass is: v₁_f = 3.66 m / s
B. Since the sign is positive, it maintains the initial direction.
C. The velocity of the heaviest mass is: v₂_f = 5.08 m / s.
D. Since the sign is positive it maintains the initial direction.
The momentum is a vector quantity defined by the product of the mass and the velocity of a body, for an isolated system there are no external forces, therefore the momentum is conserved at all instants.
They indicate that the collision is elastic, we define the system to be formed by the two bodies, therefore all the forces during the collision are internal and the momentum is conserved.
They indicate the mass of the ball 1 m₁ = 0.060 kg and its initial velocity of v₁₀ = 5.50 m / s and the mass of the ball 2 m₂ = 0.090kg moving in the same direction with velocity v₂₀ = 3.40 m / yes, let's write the momentum for two instants.
Initial instant. Before the crash.
p₀ = m₁ v₁₀ + m₂ v₂₀
Final instnate. After the crash.
p_f = m₁ v₁_ + m₂ v₂_f
The momentum is preserved.
p₀ = p_f
m₁ v₁₀ + m₂ v₂₀ = m₁ v₁_f + m₂ v₂_f
In the elastic collision the kinetic energy is conserved.
K₀ = [tex]K_f[/tex]
½ m₁ v₁₀² + ½ m₂ v₂₀² = ½ [tex]v_{1f}^2[/tex] + ½ m₂ [tex]v_{2f}^2[/tex]
Let's write our system of equations.
m₁ (v₁₀-v₁_f) = m₂ (v₂_f -v₂₀)
m₁ ( [tex]v_{1o}^2 - v_{1f}^2[/tex] ) = - m² ( [tex]v_{2f}^2 - v_{2p}^2[/tex] )
Let's use factoring in the second equation.
(a² - b²) = (a + b) (a-b)
Let's write the system
m₁ ( v₁₀-v₁_f) = m₂ ( v₂_f-v₂₀)
m₁ (v₁₀ -v₁_f) (v₁₀ + v₁_f) = m₂ (v₂_f-v₂₀) (v₂_f + v₂₀)
we solve
(v₁₀ + v₁_f) = (v₂_f + v₂₀)
We have the relations of the velocities, we substitute this equation in the expression of the conservation of momentum equation 1 and we solve.
v₁_f = v₂_f + v₂₀ -v₁₀
m₁ v₁₀ + m₂ v₂₀ = m₁ v₁_f + m₂ v₂_f
m₁ v₁₀ + m₂ v₂₀ = m₁ (v₂_f + v₂₀ -v₁₀) + m₂ v₂_f
v₁₀ (2m₁) + v₂₀ (m₂ - m₁) = v₂_f (m₁1 + m₂)
[tex]v_{2f}= \frac{2m_1}{m_1+m_2} \ v_{1o} + \frac{m_2 - m_1}{m_1+m_2} \ v_{2o}[/tex]
We solve the same for the other term.
[tex]v_{1f} = \frac{m_1-m_2}{m_1+m_2} \ v_{1o} + \frac{2m_2}{m_1+m_2} \ v_{2o}[/tex]
We substitute and calculate.
M = m₁ + m₂
M = 0.06 +0.09 = 0.150 kg
[tex]v_{2f} = \frac{2 \ 0.060 }{0.150} \ 5.50 + \frac{0.090-0.060}{0.150} \ 3.4[/tex]
v₂_f = 4.4 + 0.68
v₂_f = 5.08 m / s
[tex]v_{1f} = \frac{0.060-0.090}{0.150} \ 5.5 + \frac{2 \ 0.090 }{0.150} \ 3.4[/tex]
v₁_f = -1.1 + 4.76
v₁_f = 3.66 m / s
Having the final speeds we can answer the questions.
A. The speed of the lighter mass is: v₁_f = 3.66 m / s
B. Since the sign is positive, it maintains the initial direction.
C. The velocity of the heaviest mass is: v₂_f = 5.08 m / s.
D. Since the sign is positive it maintains the initial direction.
In conclusion using the conservation of momentum and kinetic energy we can find the results for the elastic collision of the two masses are:
A. The speed of the lighter mass is: v₁_f = 3.66 m / s
B. Since the sign is positive, it maintains the initial direction.
C. The velocity of the heaviest mass is: v₂_f = 5.08 m / s.
D. Since the sign is positive it maintains the initial direction.
Learn more about conservation of momentum and kinetic energy here: brainly.com/question/2141713
