A box initially at rest is struck by a mass of clay moving at 20 m/s. The clay sticks to a 10 kg box (initially at rest) causing it to move at 5 m/s. Calculate the mass of clay that strikes the box

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Approximately if there's no friction on the box.

Explanation:

This problem can be solved with the conservation of momentum.

The momentum of an object is equal its mass times its velocity . That is: .

In a collision between two objects, the sum of their momentum will stay the same. As long as all external forces (e.g., gravity) on the two objects are balanced, momentum will be conserved even if the two objects eventually stuck to each other.

Let the mass of the clay be .

Before the collision, the clay is moving at while the box isn't moving/

After the collision, the clay is stuck to the box. As a result, both the box and the clay will be moving at (the direction of travel is the same as that of the clay before the collision.)

If there's no friction or other unbalanced external force on the clay or the box, momentum will be conserved in the collision. As a result, .

Solve this equation for , the mass of the clay (in kilograms:) .

In other words, the mass of the clay is approximately .

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .