Answer:
= - 0.41m/s
Explanation:
Velocity of first satellite
[tex]V_1 = \frac{m_1 -m_2}{m_1 + m_2} u[/tex]
[tex]V_1 = \frac{(4 \times 10^3) - (7.5 \times 10^3)}{(4 \times 10^3) + (7.5 \times 10^3)} \times 0.15\\\\= -0.30435[/tex]
Velocity of the second satellite
[tex]V_2 = \frac{2m_1}{m_1 + m_2} u[/tex]
[tex]V_2 = \frac{2 \times 4 \times 10^3}{4 \times 10^3 + 7.5 \times 10^3} \\\\= 0.10435[/tex]
Final velocity = V(1) - V(2)
[tex]V = -0.30435 - 0.10435\\\\= -0.4087[/tex]
≅ -0.41m/s
Answer:
their final relative velocity, v₂ - v₁ = u₁ -u₂ = 0.15 m/s
Explanation:
Given;
mass of the first satellite, m₁ = 4000 kg
mass of the second satellite, m₂ = 7500 kg
their relative velocity, u₁ -u₂ = 0.15 m/s
Since, the two satellites collide elastically, then total momentum before and after collision are conserved, also the total kinetic energy is also conserved.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
m₁u₁ -m₁v₁ = m₂v₂ - m₂u₂
m₁(u₁-v₁) = m₂(v₂-u₂ ) ------------equation (i)
where;
u is the initial velocity
v is the final velocity
Also, kinetic energy is conserved
Total initial kinetic energy = Total final kinetic energy
¹/₂m₁u₁² + ¹/₂m₂u₂² = ¹/₂m₁v₁² + ¹/₂m₂v₂²
multiple through by 2
m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²
m₁u₁² - m₁v₁² = m₂v₂² - m₂u₂²
m₁(u₁² - v₁²) = m₂(v₂²- u₂²) --------------equation (ii)
divide equation (ii) by equation (i)
[tex]\frac{m_1(u_1-v_1)}{m_1(u_1^2-v_1^2)} = \frac{m_2(v_2-u_2)}{m_2(v_2^2-u_2^2)} \\\\\frac{(u_1-v_1)}{(u_1^2-v_1^2)} = \frac{(v_2-u_2)}{(v_2^2-u_2^2)}\\\\\frac{(u_1-v_1)}{(u_1-v_1)((u_1+v_1)} = \frac{(v_2-u_2)}{(v_2-u_2)(v_2+u_2)}\\\\u_1+v_1 = v_2+u_2\\\\u_1 -u_2 = v_2 -v_1[/tex]
0.15 m/s = v₂ - v₁
Thus, the relative velocity before collision is equal to the relative velocity of separation of collision.
