Help me pls Two lab carts are situated on the same horizontal, frictionless track. Cart 1 has a mass of .5 kg and initially travels to the right at 2 m/s. Cart 2 has a mass of .75 kg and initially travels to the left at 8 m/s. The two carts collide and stick together after the collision.

The law of conservation of momentum states that in a system of colliding bodies, the momentum is conserved provided there is no external force acting on the system.

Momentum of cart 1 before collision:

mass = 0.5 kg, v = 2 m/s

momentum before collision = 0.5 * 2 = 1 kgm/s

Momentum of cart 2 before collision:

mass = 0.75 kg, velocity = 8 m/s

momentum before collision = 0.75 kg * 8 m/s = 6 kgm/s

Common velocity of cart system

After collision, the stick together and move with a common velocity

From the principle of conservation of momentum;

m₁u₁ + m₂u₂ = (m₁ + m₂) * v

where v is common velocity

1 kgm/s + 6 kgm/s = (0.5 + 0.75) kg * v

v = 7/1.25 m/s

v = 5.6 m/s

Momentum of cart 1 after collision:

Momentum = 0.5 kg * 5.6 m/s

Momentum = 2.8 kgm/s

Momentum of cart 2 after collision:

Momentum = 0.75 kg * 5.6 m/s

Momentum = 4.2 kgm/s

Change in momentum of cart 1:

2.8 kgm/s - 1 kgm/s

change in momentum = 1.8 kgm/s

Change in momentum of cart 2:

4.2 kgm/s - 6 kgm/s

change in momentum = -1.8 kgm/s

Momentum of cart system

Momentum of cart system before collision = 0

Momentum after collision = (0.5 + 0.75) kg * 5.6 m/s = 7 kgm/s

Change in momentum = 7 kgm/s - 0 = 7 kgm/s

Therefore, the initial, final and change in momentum of the carts are as follows:

Cart 1: initial = 1 kgm/s; final = 2.8 kgm/s; change = 1.8 kgm/s

Cart 2: initial = 6 kgm/s; final = 4.2 kgm/s; change = 1.8 kgm/s

Cart system: initial = 0 kgm/s; final = 7 kgm/s; change = 7 kgm/s

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