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 dir

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.

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Thiên Hương 3 months 2021-07-19T22:37:19+00:00 1 Answers 4 views 0

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    2021-07-19T22:38:27+00:00

    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°

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