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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.

User Nicocesar
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1 Answer

3 votes

Answer:

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

User Mathias Stavrou
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7.3k points