Question

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.

Answer 1

Answer:

30.36°

Explanation:

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

$mv_xi + mv_yj$

where ;

m= mass

$v_x$ = velocity of components in the x direction

$v_y$ = 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 = $mvi + 2mvcos 45i + 2mvsin45 j$

= $(mv+2mvcos45)i + (2mvsin45)j$

However, the masses stick together after collision and move with a common velocity: $V_xi +V_yj$

∴ Final momentum = $3mv (V_xi +V_yj)$

= $3mV_xi + 3mV_yj$

From the foregoing ;

initial momentum = final momentum

$3mV_xi + 3mV_yj = (mv+2mvcos45)i+(2mvsin45j)$

So;

$3mV_x = mv + 2mv cos 45 \\\\3mV_y = 2mV sin 45$

$V_x = \frac{mv+2mvcos45 }{3m}\\\\V_x = \frac{v+2vcos45}{3}$

$V_y = \frac{2mvsin45}{3m} \\\\V_y = \frac{2vsin45}{3}$

Finally;

The required angle θ = $tan^{-1} = \frac{V_y}{V_x}$

θ = $tan^{-1} = \frac{\frac{2vsin45}{3}}{\frac{v+2v.cos45}{3}}$

θ = $tan^{-1} = \frac{2sin 45}{1+2cos45}$

θ = 30.36°