Final answer:
When the boy walks 10m towards the wall on the cart, due to conservation of momentum, the cart moves in the opposite direction. Since the boy is 1/4th the mass of the cart, the cart will move backwards 2.5m. Thus, the boy's final distance from the wall is 17.5m.
Step-by-step explanation:
The question describes a scenario where a 20kg boy is standing on an 80kg cart, which is initially 25m from a wall. Since there is no friction between the cart and the ground, when the boy walks towards the wall, the cart will move in the opposite direction.
This is an example of conservation of momentum where the total momentum of the system (boy plus cart) remains constant in the absence of external forces.
Let's analyze the movement: if the boy walks 10m towards the wall, due to the conservation of momentum, both the boy and the cart will move but in opposite directions.
The boy walking forward will push the cart backwards, and they must move in such a way that the product of their masses and their velocities are equal and opposite.
Since the mass of the boy is 20kg and the mass of the cart is 80kg, the boy has a mass 1/4th of that of the cart.
This means that the cart will move backwards 1/4th of the distance the boy walks forward relative to the cart.
So if the boy walks 10m relative to the cart, the cart will move 10m/4 = 2.5m backwards, away from the wall.
Therefore, the final distance of the boy from the wall will be initial distance - distance walked by the boy + distance moved by the cart which is 25m - 10m + 2.5m = 17.5m.