Question

A 144-g baseball moving 29 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 1.21 m/s .

A)Determine the baseball's speed after the collision.

B)Determine the total kinetic energy before the collision.

C)Determine the total kinetic energy after the collision.

Answer 1

Mass of the baseball m₁ = 144 g = 0.144 kg

Velocity of the baseball before collision u1 = 29 m/s

Mass of the brick m₂ = 5.25 kg

Velocity of the brick before collision u₂ = 0

Velocity of the baseball after collision be v₁

Velocity of the brick after collision be v₂ = 1.21 m/s

1. According to law of conservation of momentum,

The momentum of the system before and after collision is same

So m₁ * u₁ + 0 = m₁ * v₁ + m₂ * v₂

      0.144 * 29 = 0.144 * v₁ + 5.25 * 1.21

          4.176      = 0.144 * v₁ + 6.3525

               V₁       = ( 4.176 – 6.3525 )/0.144

                          = -2.1765/0.144

                          = - 15.11 m/s

So velocity of baseball after collision is -15.11 m/s

b ) Kinetic energy before collision

                    K.E_before = ½ * m₁ * u₁² + 0

                                   = ½ * 0.144 * ( 29 )²

                                   = 60.552 J

C ) Kinetic energy after collision

                   K.E_after    = ½ * m₁ * v₁² + ½ * m₂ * v₂²

                                  = ½ * 0.144 * (15.11)² + ½ * 5.25 * ( 1.21 )²

                                  = 16.438 + 3.843 = 20.28 J