An elastic collision is a collision in which there is no net loss in kinetic energy in the system as a result of the collision. Both momentum and kinetic energy are conserved quantities in elastic collisions. ... They collide, bouncing off each other with no loss in speed.
Answer:
0.665
Explanation:
I did the work. Just plug everything in from the formula. Look at the lesson manual.
(h + .16) m g = 1/2 k x^2 total PE of block relative to where it stops
(h + .16) .82 * 9.8 = .5 * 120 * .16^2 PE released = PE of spring
8.04 h + 1.29 = 1.536
h = (1.536 - 1.29) / 8.04 = .031 m = 3.1 cm
Answer:
1)a. It is constant the whole time the ball is in free-fall.
2)b. = 14 m/s
3) e. = 19.6 m/s
Explanation:
1) given that the only force acting on the ball is gravity, gravity acts along the vertical axis. Since no other force acts on the ball then the horizontal velocity will remain constant all through the flight since there is no horizontal force acting on the ball.
2) speed = distance/time
horizontal distance = 56m
Time = 4 seconds
Speed = 56m/4s = 14m/s
3) acceleration due to gravity g = 9.8m/s^2
Initial vertical velocity = u
Final vertical velocity = v = -u
Using the law of motion;
v = u + at
a = acceleration = -g = -9.8m/s^2
t = time of flight = 4
Substituting the values;
-u = u - 4(9.8)
-2u = -4(9.8)
u = -4(9.8)/-2
u = 2(9.8) = 19.6 m/s
Initial vertical velocity = u = 19.6 m/s
