Answer :
Answer:
The momentum before collision and momentum after collision is equal in a frame of reference moving at 10.0 m/s in the direction of the moving car.
Explanation:
[tex]m_1[/tex] = Mass of moving car = 1990 kg
[tex]u_1[/tex] = Velocity of moving car in stationary frame = 20 m/s
[tex]m_2[/tex] = Mass of stationary car = 1540 kg
[tex]u_2[/tex] = Velocity of stationary car in stationary frame = 0 m/s
v = Combined velocity in stationary frame
Momentum conservation for stationary frame
[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\dfrac{1990\times 20+0}{1990+1540}\\\Rightarrow v=\dfrac{3980}{353}\ \text{m/s}[/tex]
In frame moving at 10 m/s the velocities change in the following ways
[tex]u_1=20-10\\\Rightarrow u_1=10\ \text{m/s}[/tex]
[tex]u_2=0-10\\\Rightarrow u_1=-10\ \text{m/s}[/tex]
[tex]v=\dfrac{3980}{353}-10\\\Rightarrow v=\dfrac{450}{353}\ \text{m/s}[/tex]
Momentum before collision
[tex]m_1u_1+m_2u_2=1990\times 10+1540\times -10\\\Rightarrow m_1u_1+m_2u_2=4500\ \text{kg m/s}[/tex]
Momentum after collision
[tex](m_1+m_2)v=(1990+1540)\times \dfrac{450}{353}\\\Rightarrow (m_1+m_2)v=4500\ \text{kg m/s}[/tex]
The momentum before collision and momentum after collision is equal. So, the momentum is conserved in a reference frame moving at 10.0 m/s in the direction of the moving car.