A billiard ball moving at 570 ms strikes a stationary ball o

A billiard ball moving at 5.70 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 4.78 m/s, at an angle of 33.0° with respect to the original line of motion.

(a) Find the velocity (magnitude and direction) of the second ball after collision.
1 m/s
2 ° (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?

inelastic elastic    

Solution

a)

Initial momentum of the system, Pi = 5.7*m i <------- i = unit vector along x axis

Final momentum of the system,

Pf = m*(4.78*(cos(33 deg) i + sin(33 deg) j)) + m*v

where v = final velocity of the second ball

By conservation of momentum,

Pi = Pf

So,  5.7*m i = m*(4.78*(cos(33 deg) i + sin(33 deg) j)) + m*v

So, 5.7 i - (4.78*(cos(33 deg) i + sin(33 deg) j)) = v

So, v = 1.69i - 2.6 j

So, magnitude = sqrt(1.69^2+2.6^2) = 3.1 m/s <-------answer

direction = atan(-2.6/1.69) = -57 deg with the original line of motion

b)

Initial Kinetic energy of the system,

KEi = 0.5*m*5.7^2 = 16.25*m J

Final Kinetic energy of the system,

KEf = 0.5*m*4.78^2 + 0.5*m*3.1^2 = 16.23*n J

As KEf /= KEi , so the collision is inelastic <--------answer