A hockey player is moving northward and suddenly turns westward with the same speed to avoid an opponent. The force that acts on the player is

To determine the force acting on the hockey player, we can apply Newton's second law of motion, which states that the force acting on an object is equal to its rate of change of momentum. Since the mass of the player remains constant, the direction of the force is the same as the direction of the change in velocity.

Let us represent the directions using unit vectors:
Let the unit vector along the North direction be $\hat{j}$.
Let the unit vector along the West direction be $-\hat{i}$ (where $\hat{i}$ is along the East direction).

Initially, the player is moving northward with a speed $v$. Therefore, the initial velocity vector is:

${\overrightarrow{v}}_i=v\hat{j}$

Suddenly, the player turns westward with the same speed $v$. Therefore, the final velocity vector is:

${\overrightarrow{v}}_f=-v\hat{i}$

The change in velocity ($\Delta \overrightarrow{v}$) is given by the difference between the final velocity and the initial velocity:

$\Delta \overrightarrow{v}={\overrightarrow{v}}_f-{\overrightarrow{v}}_i$

Substituting the velocity vectors into the equation:

$\Delta \overrightarrow{v}=-v\hat{i}-v\hat{j}=-v(\hat{i}+\hat{j})$

The vector $-\hat{i}$ points West, and the vector $-\hat{j}$ points South. Therefore, the direction of the change in velocity vector $\Delta \overrightarrow{v}$ is along the South-West direction.

Since force is directly proportional to the change in velocity ($\overrightarrow{F}=m\frac{d\overrightarrow{v}}{dt}$), the force must also act in the South-West direction.

This external force is provided by the friction between the skates of the hockey player and the ice surface, allowing the player to change direction. Thus, the force acting on the player is the frictional force along the South-West direction.

Therefore, the correct option is frictional force along south-west.