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: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.