Question

A billiards ball B rests on a horizontal surface and is struck by another billiards ball A of the same mass m = 0.2 kg. Ball A is initially moving at 18 m/s and is deflected 28° from its original direction. Assume the collision is elastic. a) find the final velocity vector of each ball, b) find the mechanical energy both before and after the collision.

Answer 1

v1 = 15.90 m/s

v2 = 8.46 m/s

mechanical energy before collision = 32.4 J

mechanical energy after collision = 32.433 J

Step-by-step explanation:

given data

mass m = 0.2 kg

speed = 18 m/s

angle = 28°

to find out

final velocity and mechanical energy both before and after the collision

solution

we know that conservation of momentum remain same so in x direction

mv = mv1 cosθ + mv2cosθ

put here value

0.2(18) = 0.2 v1 cos(28) + 0.2 v2 cos(90-28)

3.6 = 0.1765 V1 + 0.09389 v2 ................1

and

in y axis

mv = mv1 sinθ - mv2sinθ

0 = 0.2 v1 sin28 - 0.2 v2 sin(90-28)

0 = 0.09389 v1 - 0.1768 v2 .......................2

from equation 1 and 2

v1 = 15.90 m/s

v2 = 8.46 m/s

so

mechanical energy before collision = 1/2 mv1² + 1/2 mv2²

mechanical energy before collision = 1/2 (0.2)(18)² + 0

mechanical energy before collision = 32.4 J

and

mechanical energy after collision = 1/2 (0.2)(15.90)² + 1/2 (0.2)(8.46)²

mechanical energy after collision = 32.433 J