Ball 1 of mass m travels at a velocity of v m/s south and collides with Ball 2 of mass 3m which is travelling at a velocity of v/2 north. If the final velocity of ball 1 is v/4 north, what is the final velocity of ball 2

2/3v m/s

Explanation:

Given that

Mass of ball 1 = m

Initial velocity u1 of ball 1 = v m/s

Mass of ball 2 = 3m

Initial velocity u2 of ball 2 = v/2 m/s

Final velocity v1 of ball 1 = v/4 m/s

Final velocity v2 of ball 2 = ?

To solve this question, we will be applying the law of conservation of momentum, which is mathematically given as

m1u1 + m2u2 = m1v1 + m2v2

If we substitute the given values directly, we have

m * v + 3m * v/2 = m * v/2 + 3m * v2

mv + 3mv/2 = mv/2 + 3m * v2, collecting like terms, we have

mv + 3mv/2 - mv/2 = 3m * v2

mv + 2mv/2 = 3m * v2

mv + mv = 3m * v2

2mv = 3m * v2, dividing both sides by 3m

v2 = 2mv/3m

v2 = 2/3v

Therefore, the final velocity v2 of ball 2 is 2/3v

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.