A billiard ball moving at 5.80 m/s strikes a stationary ball of the same mass. A

Question

A billiard ball moving at 5.80 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 4.75 m/s, at an angle of 35.0° with respect to the original line of motion.

(a) Find the velocity (magnitude and direction) of the second ball after collision.
m/s ?
° ?(with respect to the original line of motion, include the sign of your answer; consider the sign of the first ball's angle)(b) Was the collision inelastic or elastic?

inelasticelastic   

Please be as clear and detailed as possible, thank you!

Explanation / Answer

here,

let the mass of each ball be m

initial speed of 1 , u1 = 5.8 i m/s

final speed of 1 , v1 = 4.75 * (cos(35) i + sin(35) j ) m/s

v1 = 3.89 i m/s + 2.72 j m/s

let the final speed of 2 be v2

using conservation of momentum

m * u1 = m * v1 + m * v2

5.8 i = 3.89 i m/s + 2.72 j m/s + v2

v2 = 1.91 i m/s - 2.72 j m/s

a)

the magnitude of velocity of v2 , |v2| = sqrt(1.91^2 + 2.72^2) = 3.32 m/s

theta = arctan(2.72 /1.91) = 54.9 degree below the original line of motion

b)

the initial kinetic energy , KEi = 0.5 * m * u1^2

KEi = 0.5 * m * 5.8^2 = 16.8 m J

the final kinetic energy , KEf = 0.5 * m * v1^2 + 0.5 * m * v2^2

KEf = 0.5 * m * 4.75^2 + 0.5 * m * 3.32^2

KEf = 16.8 m J

as KEi = KEf

the collison is perfectly Elastic

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