Question

Two manned satellites approach one another at a relative velocity of v = 0.150 m/s, intending to dock. The first has a mass of m 1 = 4.00 × 10 3 kg and the second a mass of m 2 = 7.50 × 10 3 kg. If the two satellites collide elastically rather than dock, what is their final relative velocity?

Answer 1

Answer:

= - 0.41m/s

Explanation:

Velocity of first satellite

$V_1 = \frac{m_1 -m_2}{m_1 + m_2} u$

$V_1 = \frac{(4 \times 10^3) - (7.5 \times 10^3)}{(4 \times 10^3) + (7.5 \times 10^3)} \times 0.15\\\\= -0.30435$

Velocity of the second satellite

$V_2 = \frac{2m_1}{m_1 + m_2} u$

$V_2 = \frac{2 \times 4 \times 10^3}{4 \times 10^3 + 7.5 \times 10^3} \\\\= 0.10435$

Final velocity = V(1) - V(2)

$V = -0.30435 - 0.10435\\\\= -0.4087$

≅ -0.41m/s

Answer 2

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)

$\frac{m_1(u_1-v_1)}{m_1(u_1^2-v_1^2)} = \frac{m_2(v_2-u_2)}{m_2(v_2^2-u_2^2)} \\\\\frac{(u_1-v_1)}{(u_1^2-v_1^2)} = \frac{(v_2-u_2)}{(v_2^2-u_2^2)}\\\\\frac{(u_1-v_1)}{(u_1-v_1)((u_1+v_1)} = \frac{(v_2-u_2)}{(v_2-u_2)(v_2+u_2)}\\\\u_1+v_1 = v_2+u_2\\\\u_1 -u_2 = v_2 -v_1$

0.15 m/s = v₂ - v₁

Thus, the relative velocity before collision is equal to the relative velocity of separation of collision.