1. Consider a head-on collision between two identical billiard balls. Ball 1 is initially in motion toward ball 2, which is initially at rest. After the collision, ball 2 departs with the same velocity that ball 1 originally had. Disregard any friction between the balls and the surface. What happens to ball 1? What happens to ball 2?

first ball is at rest after collision and the kinetic energy of the first ball is transferred to the second ball after collision.

Explanation:

mass of each ball = m

initial velocity of first ball = u

initial velocity of second ball = 0 m/s

final velocity of second ball = initial velocity of first ball = u

Use conservation of momentum

let v is the final velocity of the first ball after collision

m x u + m x 0 = m x v + m x u

So, v = 0

It means after the collision the second ball moves with the velocity which is equal to the initial velocity of the first ball and the first ball comes at rest.

As there is no friction between the balls during the collision, the collision of two balls is perfectly elastic and thus the kinetic energy of the system is conserved.

The entire kinetic energy of the first ball is transferred to the second ball after the collision.

t2 equal the displacement of the particle over the same time period if t2> t/2? hint: consider a sine wave starting at t1=0, then consider a closing wave starting at t1=0, for your explanatio

answer: the direction of the wind is expressed as the direction from which the wind is blowing. for example, easterly winds blow from east to west, while westerly winds blow from west to east.

explanation: