Hello!
Recall the equation for momentum:
[tex]\huge\boxed{ p = mv}[/tex]
p = linear momentum (kgm/s)
m = mass (kg)
v = velocity (m/s)
Part 1:
We can solve for the total momentum using the above equation. Let m1 represent the 0.2 kg cart, and m2 represent the 0.4 kg cart.
[tex]p = m_1v_1' + m_2v_2'[/tex]
Since they move off together:
[tex]p = v_f(m_1 + m_2) = 0.2(0.2 + 0.4) = \boxed{0.12 kg\frac{m}{s}}[/tex]
Part 2:
Using the conservation of momentum:
[tex]m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'\\\\[/tex]
m2 was initially at rest, so:
[tex]m_1v_1 + m_2(0) = 0.12\\\\0.2(v_1) = 0.12\\\\v_1 = \frac{0.12}{0.2} = \boxed{0.6 \frac{m}{s}}[/tex]
Part 3:
We can calculate the force by first calculating the impulse exerted on the carts.
Recall the equation for impulse:
[tex]\large\boxed{I = m\Delta v = m(v_f - v_i}}[/tex]
We can use either cart, but for ease, we can use the 0.4 cart that starts from rest.
Thus:
[tex]I = 0.4(0.2 - 0) = 0.08 kg\frac{m}{s}[/tex]
Now, calculate force with the following:
[tex]\large\boxed{I = Ft, F = \frac{I}{t}}[/tex]
Plug in the values:
[tex]F = \frac{0.08}{0.02} = \boxed{4 N}[/tex]
