A 53.4 kg ice skater is at rest when she throws a 0.135 kg snowball east at 15.8 m/s. What is her velocity afterwards?

thank you in advance for your help!

Her velocity afterwards is 0.04 m/s to the west.

Explanation:

Using the conservation of momentum, we have that:

m1u1 + m2u2 = m1v1 + m2v2

Where m is the mass, u is the inicial velocity and v is the final velocity.

The inicial momentum is zero, because the ice skater and the snowball are at rest (u1 = 0 and u2 = 0)

The mass of the snowball is m1 = 0.135 kg and the final velocity is v1 = 15.8 m/s, so the momentum of the snowball is:

m1 * v1 = 0.135 * 15.8 = 2.133 kg*m/s

So from the conservation of momentum, we have that:

0 = 2.133 + m2v2

m2v2 = -2.133

The mass of the ice skater is m2 = 53.4 kg, then we have that:

53.4 * v2 = -2.133

v2 = -2.133 / 53.4 = -0.0399 m/s

Her velocity afterwards is 0.04 m/s to the west.

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 .