Rank, from largest to smallest, the following four collisions according to the magnitude of the change in the momentum of cart B, which has twice the inertia of cart A.

To rank items as equivalent, overlap them.

(1) A initially stationary, B initially moving left at 1.0m/s; stick together on impact.

(b) A initially moving right at 1.0 m/s, B initially moving left at 1.0 m/s; stick together on impact.

(3) A initially moving right at 1.0 m/s, B initially stationary; stick together on impact.

(4) A initially moving right at 1.0 m/s; after impact, A moving left at 0.33 m/s, B moving right at 0.67m/s.

The largest to smallest change in momentum with respect to magnitude of change in momentum is as follows;

1st - Collision (1) & Collision (2 or b in question)

2nd - Collision (4)

3rd - Collision (3)

Explanation:

This is because momentum is mass times into velocity. i. e.

P=m. v (kg. m/s - S. I unit)

(where p is momentum, m is mass of object and v is velocity or speed object)

If mass remains constant (real life scenario) then change in momentum is directly related to change in speed. i. e

Δp=m⋅ (Δv) = m⋅ (vf-vi) where vf is final velocity and vi is initial velocity.

By using above formula;

we can calculate change in momentum for different collisions with respect to cart B.

m = mass of cart B

Collision (1) Δp=m⋅ (Δv) = m⋅ (vf-vi) = m. (0-1.0) = -m kg. m/s (where "-" indicates deceleration or stopping of object.)

Collision (2 or b in question) Δp=m⋅ (Δv) = m⋅ (vf-vi) = m. (0-1.0) = -m kg. m/s

Collision (3) Δp=m⋅ (Δv) = m⋅ (vf-vi) = m. (0-0) = 0 (which indicates that object remains stationary before and after collision and momentum for cart B is 0)

Collision (4) Δp=m⋅ (Δv) = m⋅ (vf-vi) = m. (0.67-0) = 0.67m kg. m/s

Therefore collisions (1) and (2 or b) are ranked 1st, collision (4) ranked 2nd and collision (3) ranked 3rd.