Consider a glancing collision between two identical spheres of 0.041 kg, with on

Question

Consider a glancing collision between two identical spheres of 0.041 kg, with one of the spheres initially at rest. Initially the incoming projectile has a velocity of 1.40 m/s to the right and after the collision we observe that this sphere leaves the collision region with a velocity of 0.90 m/s at an angle of 26° to its initial direction. (Assume that the +x-axis is to the right and the +y-axis is up along the page. Angles are measured counterclockwise from the +x-axis. Do not assume this collision is elastic.)

Explanation / Answer

using momentum conservation,

0.041 x 1.40i + 0.041 x 0 = 0.041 x 0.90(cos26i + sin26 j) + 0.041v

1.40i = 0.90(cos26i + sin26 j) + v

v = 0.59i - 0.39 j m/s

speed = sqrt(0.59^2 + 0.39^2) =0.71 m/s

angle = tan-1(-0.39 / 0.59) = - 33.46 degrees

in counterclock wise direction, angle =360 - 33.46 =326.53 degrees

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