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 5.07 m/s, at an angle of 29.0 with respect to the original line of motion. (Enter the direction with respect to the original line of motion. Include the sign of your answer. Consider the sign of the first ball's angle.)

(a) Find the velocity (magnitude and direction) of the second ball after collision.
________ m/s
________ °


(b) Was the collision inelastic or elastic?
inelastic
elastic

Explanation / Answer

Law of conservation of momentum must be preserved. Assume original path of motion is positive x direction mass of billard balls are same, so law of conservation of momentum can be thought of as law of conservation of velocity since m1v1i + m2v2i = m1v1f + m2v2f m1 = m2 mv1i + mv2i = mv1f + mv2f cancel m from both sides v1i + v2i = v1f + v2f initial horizontal velocity is 5.80 initial vertical velocity is 0 velocity of one ball is given at 5.07 m/s at 29.0 degrees. Vx = 5.07cos29 = 4.43 Vy = 5.07sin29 = 2.46 to sum to initial values, Vx of ball2 = 5.80 - 4.43 = 1.37 Vy of ball 2 = 0 - 2.46 = -2.46 V total =sqrt (1.37^2 + 2.46^2) =2.81 m/s angle =tan^-1(2.46/1.37) =60.9 degrees in opposite direction of initial path of movement b) initial kinetic energy =(1/2)mv^2 =16.82m final kinetic energy of ball 1 =(1/2)mv^2 =12.85m final kinetic energy of ball 2 =(1/2)mv^2 =3.95m for elastic collision, initial kinetic energy must equal final kinetic energy KEi =16.82m J KEf =12.85m J + 3.95m J =16.8m J Even though there is slight deviation, error is only 0.084%, so this collision can be thought of as elastic. Summary a) 2.81 m/s, 60.9 degrees in opposite direction of ball 1 b) elastic Hope it helps :D

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