Question

A ball of mass 1 kg dropped from 20m height on ground and it rebounds to height 5 find change in momentum during collision

Answer 1

Answer:

The change in momentum is 30 kgm/s

Explanation:

A ball of mass 1 kg dropped from 20 m height on ground and it rebounds to height 5.

Using energy conservation theorem:

$mgh=\dfrac{1}{2}mv^2$

$v=\sqrt{2gh}$

where, g = 10 m/s² and h = 20 m

$v=\sqrt{2\cdot 10\cdot 20}$

$v= 20\ m/s$

Momentum before collision = mv

                                              = 1 x 20

                                             = 20 kgm/s

After rebounding the ball.

Using energy conservation theorem:

$mgh=\dfrac{1}{2}mv^2$

$v=\sqrt{2gh}$

where, g = 10 m/s² and h = 5 m

$v=-\sqrt{2\cdot 10\cdot 5}$

$v= -10\ m/s$

Momentum after collision = mv

                                              = -1 x 10

                                             = -10 kgm/s

Change in momentum during collision = After collision - Before collision

                                                                    = 20 - (-10)

                                                                    = 30 kgm/s

Hence, The change in momentum is 30 kgm/s

Answer 2

Answer:

30 kg m/s answer is given in the attachment