Question #157279

A particle of mass m moving at 5.0m/s in the positive x direction makes a glancing elastic collision with a particle of mass 2m that is at rest before the collision. After the collision, m moves off at an angle of 450 to the x axis and 2m moves off 600 to the x axis. Calculate the speed of m after the collision.Â Â

Expert's answer

Let "v_1 = 5 \\frac{m}{s}" be the initial speed of the particle of mass m.

Let "v_1'" and "v_2'" be the speed of particle of mass "m" and "2m" after the collision respectively.

Hence, the law of conservation of momentum "\\bold p_1 = \\bold p_1' + \\bold p_2'" , written in coordinate form is:

"(m v_1, 0) = (m v_1' \\cos 45, m v_1' \\sin 45) + (2 m v_2' \\cos 60, -2 m v_2' \\sin 60)".

From y components of the last equation, obtain "v_2' = \\frac{v_1' \\sin 45}{2 \\sin 60}".

From x components of the last equation, obtain: "v_1' = \\frac{v_1 - 2 v_2' \\cos 60}{\\cos 45}" .

Substituting "v_2'" from the first equation into the second, obtain:

"v_1' = \\frac{v_1 - v_1' \\frac{\\sin 45 \\cos 60}{\\sin 60}}{\\cos 45} = \\frac{v_1}{\\cos 45} - \\tg 45 \\ctg 60 v_1'", from where "v_1' = \\frac{v_1}{\\cos 45 (1 + \\tg 45 \\ctg 60)} \\approx 4.48 \\frac{m}{s}".

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