Question

6. A 145-g baseball moving 30.5 m/s strikes a stationary 5.75-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 1.10 m/s. a. What is the baseball’s speed after the collision? (10pts) b. Find the total kinetic energy before the collision. (10pts) c. Find the total kinetic energy after the collision. (10pts)

Answer 1

Step-by-step explanation:

a )

momentum of baseball before collision

mass x velocity

= .145 x 30.5

= 4.4225 kg m /s

momentum of brick after collision

= 5.75 x 1.1

= 6.325 kg m/s

Applying conservation of momentum

4.4225 + 0 = .145 x v + 6.325 , v is velocity of baseball after collision.

v = - 13.12 m / s

b )

kinetic energy of baseball before collision = 1/2 mv²

= .5 x .145 x 30.5²

= 67.44 J

Total kinetic energy before collision = 67.44 J

c )

kinetic energy of baseball after collision = 1/2 x .145 x 13.12²

= 12.48 J .

kinetic energy of brick after collision

= .5 x 5.75 x 1.1²

= 3.48 J

Total kinetic energy after collision

= 15.96 J