A boy weighing 50 kg runs at a speed of 3 km / h behind a cart, the mass of which is 3 kg, and the speed is 2 km / h

A boy weighing 50 kg runs at a speed of 3 km / h behind a cart, the mass of which is 3 kg, and the speed is 2 km / h. The boy catches up and jumps into the cart. Determine the speed of the boy along with the cart

We translate the values ​​from given to SI:
v1 = 3 km / h = 3 * 1000/3600 = 0.83 m / s
v2 = 2 km / h = 2 * 1000/3600 = 0.56 m / s
Law of conservation of momentum: the momentum of a system remains constant for any interactions within the system.
P1 + P2 = P
Since all directions of movement are in one direction, then when projected onto the x-axis, the signs will not change.
m1 * v1 + m2 * v2 = (m1 + m2) * v
Where m1 = the mass of the boy, v1 is the speed of the boy, m2 is the mass of the cart, v2 is the speed of the cart, v is the speed of the cart with the boy.
Let us express the speed of the cart with the boy from this expression:
v = (m1 * v1 + m2 * v2) / (m1 + m2)
Substituting the numerical values, we get:
v = (m1 * v1 + m2 * v2) / (m1 + m2) = v = (50 * 0.83 + 3 * 0.56) / (50 + 3) = 0.81 m / s.
Answer: the speed of the cart with the boy is 0.81 m / s.