Question #106072

A stationary ball, with a mass of 0.5 kg, is struck by an identical ball moving at

6.0ms .

1

After the collision, the second ball moves 30° to the left of its original

direction. The stationary ball moves 60° to the right of the moving ball’s original

direction. What is the velocity of each ball after the collision?

6.0ms .

1

After the collision, the second ball moves 30° to the left of its original

direction. The stationary ball moves 60° to the right of the moving ball’s original

direction. What is the velocity of each ball after the collision?

Expert's answer

"\\overrightarrow{v}" - velocity of the second ball before the collision

"|\\overrightarrow{v}|=6\\frac{m}{s}"

"\\overrightarrow{v_1},\\overrightarrow{v_2}" - velocities of the first and second balls after the collision accordingly

"m" - the mass of each ball

According to the law of impulse conservation,

"m\\overrightarrow{v}=m\\overrightarrow{v_1}+m\\overrightarrow{v_2}\\rArr \\overrightarrow{v}=\\overrightarrow{v_1}+\\overrightarrow{v_2}"

So "\\overrightarrow{v}" is the summ of vectors "\\overrightarrow{v_1}" and "\\overrightarrow{v_2} \\rArr" according to the triangle law of adding vectors, "\\overrightarrow{v}, \\overrightarrow{v_1}, \\overrightarrow{v_2}" form a triangle

The angle between "\\overrightarrow{v_1}" and "\\overrightarrow{v_2}" is equal to "30^o+60^o=90^o"

So this triangle is right "\\rArr |\\overrightarrow{v_1}|=|\\overrightarrow{v}|cos60^o=6\\frac{m}{s}*\\frac{1}{2}=3\\frac{m}{s}"

"|\\overrightarrow{v_2}|=|\\overrightarrow{v}|cos30^o=6\\frac{m}{s}*\\frac{\\sqrt{3}}{2}=3\\sqrt{3}\\frac{m}{s}\\approx5.2\\frac{m}{s}"

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