Question

A 1990 kg car moving at 20.0 m/s collides and locks together with a 1540 kg car at rest at a stop sign. Show that momentum is conserved in a reference frame moving at 10.0 m/s in the direction of the moving car.

Answer 1

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.

Step-by-step explanation:

`m_1` = Mass of moving car = 1990 kg

`u_1` = Velocity of moving car in stationary frame = 20 m/s

`m_2` = Mass of stationary car = 1540 kg

`u_2` = Velocity of stationary car in stationary frame = 0 m/s

v = Combined velocity in stationary frame

Momentum conservation for stationary frame

`m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=(m_1u_1+m_2u_2)/(m_1+m_2)\\\Rightarrow v=(1990* 20+0)/(1990+1540)\\\Rightarrow v=(3980)/(353)\ \text{m/s}`

In frame moving at 10 m/s the velocities change in the following ways

`u_1=20-10\\\Rightarrow u_1=10\ \text{m/s}`

`u_2=0-10\\\Rightarrow u_1=-10\ \text{m/s}`

`v=(3980)/(353)-10\\\Rightarrow v=(450)/(353)\ \text{m/s}`

Momentum before collision

`m_1u_1+m_2u_2=1990* 10+1540* -10\\\Rightarrow m_1u_1+m_2u_2=4500\ \text{kg m/s}`

Momentum after collision

`(m_1+m_2)v=(1990+1540)* (450)/(353)\\\Rightarrow (m_1+m_2)v=4500\ \text{kg m/s}`

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.