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A 144-g baseball moving 27 m/s strikes a stationary 5.25-kg brick resting on small rollers so it moves without significant friction. After hitting the brick, the baseball bounces straight back, and the brick moves forward at 1.07 m/s .

1 Answer

3 votes

Answer:

A. - 12.01 m/s

B. Kinetic energy before collision = 52.488 J

Kinetic energy after collision = 13.40 J

Step-by-step explanation:

Parameters given:

Mass of ball = 144g = 0.144kg

Initial velocity of ball = 27 m/s

Mass of brick = 5.25kg

Initial velocity of brick = 0 m/s

Final velocity of brick = 1.07m/s

A. To find the Final velocity of the ball, we use the law of conservation of momentum i.e.

Total initial momentum = Total final momentum

Initial momentum of ball = m*u

= 0.144 * 27

= 3.888 kgm/s

Initial momentum of brick = m*u

= 5.25 * 0

= 0 kgm/s

Total initial momentum = 3.888 + 0

= 3.888 kgm/s

Final momentum of ball = 0.144 * v

= 0.144v

Final momentum of brick = 5.25 * 1.07

= 5.6175 kgm/s

Total final momentum = 0.144v + 5.6175

Hence,

3.888 = 0.144v + 5.6175

=> 0.144v = 3.888 - 5 6175

0.144v = -1.7295

v = - 1.7295/0.144

v = - 12.01m/s

Therefore, the final velocity of the ball is - 12.01m/s. The negative sign indicates that it's moving opposite to the initial velocity.

B. Kinetic energy of ball before collision =

0.5 * m * u^2

= 0.5 * 0.144 * 27^2

= 52.488 J

Kinetic energy of the brick before collision =

0.5 * 5.25 * 0

= 0 J

Total kinetic energy before collision is:

52.488 + 0 = 52.488 J

Kinetic energy of ball after collision

= 0.5 * 0.144 * (12.01)^2

= 10.39J

Kinetic energy of brick after collision

= 0.5 * 5.25 * 1.07^2

= 3.01 J

Total kinetic energy after collision

= 10.39 + 3.01

= 13.40 J

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