Question

an object of mass 2 kg is
dropped from the height of
50m. If it experiences an
acceleration Of 10m/s2, find
the speed by which it hits the
ground. If the objects rebounds
with the speed of 9 m/s then
calculate the change in
momentum​

Answer 1

Answer:

20kgm/s - (-20kgm/s)

40kgm/s in upward

Explanation:

mass, m = 1kg

height, s = 20m

initial velocity of ball, u = 0m/s

acceleration, a = 10m/s2

Using, v2 = u2 + 2as

                = 0 + 2×10 ×20

                = 400

             v = 20m/s

Now, let us take upward direction as positive and downward direction as negative.

Initial momentum of the ball (before striking the ground) = m×v

                                                                                              = 1×(-20)  (velocity is downward, hence negative)

                                                                                              = -20kgm/s 

Now, when the ball strikes and rebounds, its velocity becomes +20m/s (after striking, it moves upward, hence positive velocity)

Final momentum (after striking the ground) = m×v

                                                                         = 1×(+20)

                                                                         = +20kgm/s

Change in momentum = Final momentum - Initial momentum

                                      = 20kgm/s - (-20kgm/s)

                                      = 40kgm/s in upward direction.

Answer 2

acceleration = velocity/time=10/2 velocity is 10 time is 2

velocity =speed/time

hence, speed = velocity*time = 10*2

speed = 20m/s