Question

a body of mass 2kg is dropped from a height of 200 metres. what is it's momentum just before it touches the ground ?

can anyone please answer this​

Answer 1

Answer:

mass, m = 1kg

height, s = 20m

initial velocity of ball, u = 0m/s

acceleration, a = 10m/s

2

Using, v

2

= u

2

+ 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

Answer:

like this you can solve

Explanation:

Velocity of 1 kg block just before it collides with 3kg block =2gh−−−√=2000−−−−√m/s

Applying momentum conversation just before and just after collision.

1×2000−−−−√=4v⇒v=2000√4m/s

initial compression of spring

1.25×106x0=30⇒x0≈0

applying work energy theorem,

Wg+Wsp=ΔKE

⇒40×x+12×1.25×106(02−x2)

=0−12×4×v2

solving x≈4 cm