On a frictionless horizontal air table, puck A (with mass 0.255 kg ) is moving toward puck B (with mass 0.365 kg ), which is initially at rest. After the collision, puck A has velocity 0.118 m/s to the left, and puck B has velocity 0.649 m/s to the right.

What was the speed vAi of puck A before the collision?

This is the same as the one with the skaters, just different numbers

momentum before = .255 v + .365 * 0
momentum after = .255 (-.118) + .365*.649

momentum after = momentum before, solve for v

To find the speed of puck A before the collision (vAi), we can use the principles of conservation of momentum.

The law of conservation of momentum states that the total momentum of a system before a collision is equal to the total momentum after the collision, provided there are no external forces acting on the system.

Mathematically, we can express this as:

m1 * v1i + m2 * v2i = m1 * v1f + m2 * v2f

Where:
m1, m2 are the masses of the pucks A and B respectively,
v1i, v2i are the initial velocities of pucks A and B respectively,
v1f, v2f are the final velocities of pucks A and B respectively.

Since puck B is initially at rest (v2i = 0), we can simplify the equation to:

m1 * v1i = m1 * v1f + m2 * v2f

Now, we can substitute the given values into the equation:

0.255 kg * vAi = 0.255 kg * 0.118 m/s + 0.365 kg * 0.649 m/s

Simplifying further, we have:

0.255 kg * vAi = 0.03009 kg*m/s + 0.23685 kg*m/s

Combining the terms on the right side:

0.255 kg * vAi = 0.26694 kg*m/s

Finally, solving for vAi:

vAi = (0.26694 kg*m/s) / 0.255 kg
vAi ≈ 1.048 m/s

Therefore, the speed of puck A (vAi) before the collision is approximately 1.048 m/s.