Question

A billiard ball (m=0.17 kg) traveling at 3.25 m/s collides with an identical billiard ball at rest. After the collision, the first ball travels at 1.87 m/s at an angle of 23 degrees from the positive x-axis. What is (a) the speed and (b) the angle of the second ball?

Answer #1

here,

mass of each ball , m = 0.17 kg

initial speed of ball 1 , u1 = 3.25 m/s

the final speed of ball 1 , v1 = 1.87 m/s

theta1 = 23 degree

let the velocity of 2 after the collison be v2

using conservation of momentum

m * u1 = m * v1 * ( cos(theta1) i + sin(theta1) j) + m * v2

3.25 i = 1.8 * ( cos(23) i + sin(23) j) + v2

v2 = 1.59 i m/s - 0.703 j m/s

a)

the speed of 2 , |v2| = sqrt(1.59^2 + 0.703^2)

**|v2| = 1.74 m/s**

b)

the angle of seccond ball , theta2 = arctan(0.703/1.59)

**theta2 = 336 degree counterclockwise from +x
axis**

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