A7.0-kilogram cart, a, and a 3.0-kilogram cart, b, are initially held together at rest on a horizontal, frictionless surface. when a compressed spring attached to one of the carts is released, the carts are pushed apart. after the spring is released, the speed of cart b is 6.0 meters per second, as represented in the diagram belowwhat is the speed of cart a after the spring is released?

For this problem, we use the conservation of momentum as a solution. Since momentum is mass times velocity, then,

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
where
v₁ and v₂ are initial velocities of cart A and B, respectively
v₁' and v₂' are final velocities of cart A and B, respectively
m₁ and m₂ are masses of cart A and B, respectively

(7 kg)(0 m/s) + (3 kg)(0 m/s) = (7 kg)(v₁') + (3 kg)(6 m/s)
Solving for v₁',
v₁' = -2.57 m/s

Therefore, the speed of cart A is at 2.57 m/s at the direction opposite of cart B.

The speed of cart A after the spring is released is 2.57 m/s.

Given the following data:

Mass of cart A = 7.0 kgInitial velocity of cart A and B = 0 m/s (since they were held at rest).Mass of cart B = 3.0 kgFinal velocity of cart B = 6.0 m/s

To find the speed of cart A after the spring is released, we would apply the law of conservation of momentum:

Where:

is the mass of the first cart. is the mass of the second cart. are the initial velocities. is the initial velocity of the first cart. is the final velocity of the second cart.

Substituting the given parameters into the formula, we have;

Note: The negative sign indicate that cart A is moving to the left direction.

Wish i knew this to !