You can view this as an inelastic collision between a 2000 kg object moving at 3 m/s, and a stationary 500 kg object.
The total momentum of the objects before the collision equals their momentum after collision.
Before:
flatcar = 2000*3 = 6000 kg*m/s.
<span> coal = 0*500 = 0.
After:
flatcar & coal = 2500*V
then
</span> 2500*V <span>= 6000,
V = 2.4 m/s.
</span> Additional:
the definition of an inelastic collision is that the objects "stick together" after colliding.
consider the motion of the ball in vertical direction after it rolls off the table
v₀ = initial velocity of the ball in vertical direction = 0 m/s
a = acceleration = acceleration due to gravity = 9.8 m/s²
Y = vertical displacement = height of the table = 0.91 m
t = time spent in air by the ball
Using the kinematics equation
Y = v₀ t + (0.5) a t²
0.91 = 0 t + (0.5) (9.8) t²
t = 0.43 sec
v = final velocity in vertical direction just before it hits the ground
Using the kinematics equation
v = v₀ + at
v = 0 + (9.8) (0.43)
v = 4.21 m/s
Try to split the questions for easy access
