All Balls are identical so have same mass M.
P x i = 5 M P x f = 4.33 cos 30 ° M + V sin θ M P_{xi}=5M\\
P_{xf}=4.33\cos{30\degree}M+ V\sin{\theta}M\\ P x i = 5 M P x f = 4.33 cos 30° M + V sin θ M
Applying conservation of linear momentum along original line of motion.
V cos θ = 5 − 4.33 × 3 2 . . . . . . E q [ 1 ] V\cos{\theta}=5-4.33\times{\cfrac{\sqrt{3}}{2}}......Eq[1] V cos θ = 5 − 4.33 × 2 3 ...... Eq [ 1 ] V cos θ = 5 − 4.33 × 3 2 . . . . . . E q [ 1 ] V\cos{\theta}=5-4.33\times{\cfrac{\sqrt{3}}{2}}......Eq[1] V cos θ = 5 − 4.33 × 2 3 ...... Eq [ 1 ] Applying conservation of linear momentum along perpendicular direction.
P y i = P y f 0 = 4.33 M sin 30 ° − M V sin θ V sin θ = 4.33 2 . . . . . . E q [ 2 ] P_{yi}=P_{yf}\\
0=4.33M\sin{30\degree}-MV\sin{\theta}\\
V\sin{\theta}= \cfrac{4.33}{2}......Eq[2]\\ P y i = P y f 0 = 4.33 M sin 30° − M V sin θ V sin θ = 2 4.33 ...... Eq [ 2 ] on solving Eq[1] & Eq[2], we get,
V = ( 5 − 4.33 3 2 ) 2 + ( 4.33 2 ) 2 V = 2.5 m / s e c V=\sqrt{\bigg(5-4.33\cfrac{\sqrt{3}}{2}\bigg)^2+\bigg(\cfrac{4.33}{2}\bigg)^2}\\
V=2.5m/sec V = ( 5 − 4.33 2 3 ) 2 + ( 2 4.33 ) 2 V = 2.5 m / sec V = ( 5 − 4.33 3 2 ) 2 + ( 4.33 2 ) 2 V = 2.5 m / s e c V=\sqrt{\bigg(5-4.33\cfrac{\sqrt{3}}{2}\bigg)^2+\bigg(\cfrac{4.33}{2}\bigg)^2}\\
V=2.5m/sec V = ( 5 − 4.33 2 3 ) 2 + ( 2 4.33 ) 2 V = 2.5 m / sec V = ( 5 − 4.33 3 2 ) 2 + ( 4.33 2 ) 2 V = 2.5 m / s e c V=\sqrt{\bigg(5-4.33\cfrac{\sqrt{3}}{2}\bigg)^2+\bigg(\cfrac{4.33}{2}\bigg)^2}\\
V=2.5m/sec V = ( 5 − 4.33 2 3 ) 2 + ( 2 4.33 ) 2 V = 2.5 m / sec and
θ = tan − 1 ( 4.33 10 − 4.33 × 3 ) θ = tan − 1 ( 1.72 ) = 59.8 ° ≈ 60 ° \theta=\tan^{-1}\bigg(\cfrac{4.33}{10-4.33\times{\sqrt{3}}}\bigg)\\
\theta=\tan^{-1}(1.72)=59.8\degree\approx60\degree θ = tan − 1 ( 10 − 4.33 × 3 4.33 ) θ = tan − 1 ( 1.72 ) = 59.8° ≈ 60° θ = tan − 1 ( 4.33 10 − 4.33 × 3 ) θ = tan − 1 ( 1.72 ) = 59.8 ° ≈ 60 ° \theta=\tan^{-1}\bigg(\cfrac{4.33}{10-4.33\times{\sqrt{3}}}\bigg)\\
\theta=\tan^{-1}(1.72)=59.8\degree\approx60\degree θ = tan − 1 ( 10 − 4.33 × 3 4.33 ) θ = tan − 1 ( 1.72 ) = 59.8° ≈ 60° yi is the same as yf so the collision was perfectly elastic since it is consistent with other data
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