An elastic collision of two pucks (mA = 0.5 kg, mg = 0.3 kg) on a frictionless air-jockey table. Puck A has an initial velocity of 4m/s in the positive X-direction and a final velocity of 2m/s in an unknown direction a. Puck B is initially at rest. Find the final speed of puck B and the angles a and b.
For an elastic collision
X: Mv=mu′⋅cosβ+Mv′⋅cosαMv=mu'\cdot \cos\beta+Mv'\cdot\cos\alphaMv=mu′⋅cosβ+Mv′⋅cosα
Y: mu′⋅sinβ=Mv′⋅sinαmu'\cdot \sin\beta=Mv'\cdot\sin\alphamu′⋅sinβ=Mv′⋅sinα
KE: Mv2/2=mu′2/2+Mv′2/2→Mv^2/2=mu'^2/2+Mv'^2/2 \toMv2/2=mu′2/2+Mv′2/2→
0.5⋅42/2=0.3⋅u′2/2+0.5⋅22/2→u′=4.47(m/s)0.5\cdot4^2/2=0.3\cdot u'^2/2+0.5\cdot 2^2/2\to u'=4.47(m/s)0.5⋅42/2=0.3⋅u′2/2+0.5⋅22/2→u′=4.47(m/s) .Answer
0.5⋅4=0.3⋅4.47⋅cosβ+0.5⋅2⋅cosα0.5\cdot 4=0.3\cdot 4.47\cdot \cos\beta+0.5\cdot 2\cdot\cos\alpha0.5⋅4=0.3⋅4.47⋅cosβ+0.5⋅2⋅cosα
0.3⋅4.47⋅sinβ=0.5⋅2⋅sinα→0.3\cdot 4.47\cdot \sin\beta=0.5\cdot 2\cdot\sin\alpha\to0.3⋅4.47⋅sinβ=0.5⋅2⋅sinα→ α≈36.99°\alpha\approx36.99°α≈36.99° and β≈26.66°\beta\approx26.66°β≈26.66° . Answer
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