Answer to Question #142722 in Mechanics | Relativity for Fletcher Madden

Question #142722
A billiard ball moving at 5.00 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 4.33 m/s at an angle of 30.0degrees with respect to the original line of motion. Assuming an elastic collision, find the struck ball's velocity and direction after the collision.
1
Expert's answer
2020-11-30T14:59:20-0500

All Balls are identical so have same mass M.


Pxi=5MPxf=4.33cos30°M+VsinθMP_{xi}=5M\\ P_{xf}=4.33\cos{30\degree}M+ V\sin{\theta}M\\


Applying conservation of linear momentum along original line of motion.

Vcosθ=54.33×32......Eq[1]V\cos{\theta}=5-4.33\times{\cfrac{\sqrt{3}}{2}}......Eq[1]

Applying conservation of linear momentum along perpendicular direction.

Pyi=Pyf0=4.33Msin30°MVsinθVsinθ=4.332......Eq[2]P_{yi}=P_{yf}\\ 0=4.33M\sin{30\degree}-MV\sin{\theta}\\ V\sin{\theta}= \cfrac{4.33}{2}......Eq[2]\\

on solving Eq[1] & Eq[2], we get,

V=(54.3332)2+(4.332)2V=2.5m/secV=\sqrt{\bigg(5-4.33\cfrac{\sqrt{3}}{2}\bigg)^2+\bigg(\cfrac{4.33}{2}\bigg)^2}\\ V=2.5m/sec

and

θ=tan1(4.33104.33×3)θ=tan1(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

Hence, Velocity of second ball is 2.5 m/sec and its direction from original line of motion is 600.

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