A ball has a mass of 4.65 kg and approximates a ping pong ball of mass 0.060 kg that is at rest by striking it in an elastic collision. The initial velocity of the bowling ball is 5.00 m / s, determine the final velocities of both masses after the collision. Use equations 9.21 and 9.22 from the textbook. The book is on WebAssign.

the final velocity of the ball is 4.87 m/s

the final velocity of the ping ball is 9.87 m/s

Explanation:

Given;

mass of the ball, m₁ = 4.65 kg

mass of the ping ball, m₂ = 0.06 kg

initial velocity of the ping ball, u₂ = 0

initial velocity of the ball, u₁ = 5 m/s

let the final velocity of the ball = v₁

let the final velocity of the ping ball, = v₂

Apply the principle of conservation of linear momentum for elastic collision;

m₁u₁  +  m₂u₂  = m₁v₁   +  m₂v₂

4.65(5)  +   0.06(0)   =   4.65v₁   +   0.06v₂

23.25 + 0 = 4.65v₁  +  0.06v₂

23.25 = 4.65v₁  +  0.06v₂   (1)

Apply one-directional velocity equation;

u₁ + v₁ = u₂  +  v₂

5 + v₁ = 0  +  v₂

5 + v₁ = v₂

v₁ = v₂ - 5   (2)

substitute equation (2) into (1)

23.25 = 4.65(v₂ - 5)  +  0.06v₂

23.25 = 4.65v₂  -  23.25   +   0.06v₂

46.5 = 4.71 v₂

v₂ = 46.5/4.71

v₂ = 9.87 m/s

v₁ = v₂ - 5

v₁ = 9.87 - 5

v₁ = 4.87 m/s

answer: option c.

explanation:

= initial pressure

= final pressure = 2

= initial volume

= final volume = /4

= initial temperature = 20 c = 20 + 273 = 293 k

= final temperature = ?

using the equation