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3 years ago

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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 co

nserved in a reference frame moving at 10.0 m/s in the direction of the moving car.

1 answer:

Ilya [14]3 years ago

7 0

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:

$m_1$ = 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

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

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

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

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

$v=\dfrac{3980}{353}-10\\\Rightarrow v=\dfrac{450}{353}\ \text{m/s}$

Momentum before collision

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

Momentum after collision

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

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.

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