A block of mass m moving due east at speed v collides with and sticks to a block of mass 2m that is moving at the same speed vv but in a direction 45.0∘ north of east. Find the direction in which the two blocks move after the collision. Express your answer as an angle θ in degrees measured north of east.

30.36°

Explanation:

By using linear momentum; linear momentum can be expressed by the relation:

where ;

m= mass

= velocity of components in the x direction

= velocity of components in the y direction

If we consider the east as the positive x and north as positive y which is synonymous to what we usually have on a graph.

Then;

Initial momentum =

=

However, the masses stick together after collision and move with a common velocity:

∴ Final momentum =

=

From the foregoing ;

initial momentum = final momentum

So;

Finally;

The required angle θ =

θ =

θ =

θ = 30.36°

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