Work Energy Theorem

We have , so .

This is not constant acceleration, so we need to use a different approach. Let’s try to rewrite the equation:

So,

Integrating both sides with respect to :

At , and , so: =>

Thus, and

Now, , so

To solve , complete the square in the denominator:

Let , then

So,

Thus,

Since , then and

Therefore,

At , , so

So,

At : m

At : m/s

Momentum kg m/s. Note that the problem states that the mass move ALONG the x direction, so we should keep the momentum positive.

Answer: (C) m, kg m/s