Question

How fast must a 2.5-g ping-pong ball move in order to have the same kinetic energy as a 145 g baseball moving at 38.0 m/s?

Answer 1

Kinetic energy = (1/2)*m*v^2. So we have (1/2)*(145 g)*(37 m/s)^2 = (1/2)*(2.5 g)*(v^2).

Solve for v = 281.78 m/s {about 630 miles per hour}

Step-by-step explanation:

Answer 2

Velocity of ping-pong ball = V1 = 289.4 m/s

Step-by-step explanation:

Mass of ping-pong ball = m1 = 2.5 g = 0.0025 kg

Mass of baseball = m2 = 145 g = 0.145 kg

Velocity of baseball = V2 = 38 m/s

Velocity of ping-pong ball = V1 = ?

Kinetic energy of baseball = K.E2 = (1/2)(m2)(V2)^2

= (1/2)(0.145)(38)^2

= 104.69 J

It is given that:

Kinetic of ping-pong ball = kinetic energy of baseball

K.E1 = K.E2

(1/2)(m1)(V1)^2 = 104.69

V1 = √(104.69)(2)/0.0025

V1 = 289.4 m/s