A 0.50 kg ball that is traveling at 9.0 m/s collides head-on with a 1.00 kg ball moving in the opposite direction at a speed of 11.0 m/s. The 0.50 kg ball bounces backward at 15.0 m/s after the collision. Find the speed of the second ball after the collision.

1 m/s

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu + m'u' = mv + m'v' Equation 1

Where m = mass of the first ball m' = mass of the second ball, u = initial velocity of the first ball, u' = initial velocity of the second ball, v = final velocity of the first ball, v' = final velocity of the second ball.

Note: Let the direction of the first ball be positive.

Given: m = 0.5 kg, u = 9.0 m/s, m' = 1 kg, u' = - 11 m/s ( opposite direction to the first ball), v = - 15 m/s ( bounces backward).

Substitute into equation 1

0.5(9) + 1(-11) = 0.5(-15) + 1(v')

Solving for v'

4.5-11 = -7.5 + v'

-6.5 = -7.5 + v'

v' = 7.5-6.5

v' = 1 m/s.

Hence the speed of the second ball after collision = 1 m/s

falling water hydro = water .. spectcular places these plants are

massxgravity x heght=weightxheight= pe

4900=490xh

h=10