A billiard ball moving at 5.40 m/s strikes a stationary ball of thesame mass. Af

Question

A billiard ball moving at 5.40 m/s strikes a stationary ball of thesame mass. After the collision, the first ball moves at4.81 m/s, at an angle of 27.0° withrespect to the original line of motion. (a) Find the velocity (magnitude and direction) of the second ballafter collision.
m/s
°(with respect to the original line of motion, include the sign ofyour answer; consider the sign of the first ball's angle)
(a) Find the velocity (magnitude and direction) of the second ballafter collision.
m/s
°(with respect to the original line of motion, include the sign ofyour answer; consider the sign of the first ball's angle)

Explanation / Answer

Let the mass of the ball be m before collision : ---------------- speed of the 1 st ball u = 5.4 m / s speed of the 2 nd ball U = 0 m / s after collision : -------------- speed of 1 st ball v = 4.81 m / s angle = 27 degrees Let the speed of the 2 nd ball be V In original line of motion : ------------------------ from law of conservation of momentum, m u = m v cos 27 +m V cos                                                            5.4 = 4.81 cos 27 + V cos                ----------(1) Perpendicular to the original direction : -------------------------------------- from law of conservation of momentum , 0 = m v sin 27 + m Vsin from this   V sin = - v sin 27                            = - 4.81 sin 27                           = 2.183 -----------------(2) eq( 2 ) / eq ( 1 ) ==> we get    value substitue this val;ue in eq ( 1) weget   V   value

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