Answer:
The speed of the 270g cart after the collision is 0.68m/s
Explanation:
Mass of air track cart (m1) = 320g
Initial velocity (u1) = 1.25m/s
Mass of stationary cart (m2) = 270g
Velocity after collision (V) = m1u1/(m1+m2) = 320×1.25/(320+270) = 400/590 = 0.68m/s
Answer:
Part a)

Part b)

Part c)

Explanation:
Part a)
As we know that the friction force on two boxes is given as



Now we know by Newton's II law

so we have




Part b)
For block B we know that net force on it will push it forward with same acceleration so we have




Part c)
If Alex push from other side then also the acceleration will be same
So for box B we can say that Net force is given as




Basically it is the difference in velocity divided by the time it takes to make that change.