How much momentum will an object of mass 10 kg transfer to the floor i

1.	How much momentum will an object of mass 10 kg transfer to the floor if it falls from a height of 5 m? (g=10m/s)
Answer» Answer: Given: MASS = 10 kg Height = 5 metres To FIND: Momentum transferred to the ground. Concept: The final momentum of the object (just before hitting the ground) will be transferred to the ground, if the collision is considered elastic. Calculation: Since the object is dropped, initial velocity of object is zero. ${v}^{2} = {u}^{2} + 2gh \\ = > {v}^{2} = {0}^{2} + (2 \times 10 \times 5) \\ = > {v}^{2} = 100 \\ = > v = \sqrt{100 } \\ = > v = 10 \: m {s}^{ - 1}$ So, final momentum of object just before hitting the ground . $\therefore \: p1 = m \times v \\ = > p1 = 10 \times 10 \\ = > p1 = 100 \: kg \: m {s}^{ - 1}$ Momentum after hitting the ground : $p2 = 0 \: kg \: m {s}^{ - 1}$ HENCE, momentum CHANGE of the object is correspondingly transferred to the ground. So , momentum sent to ground $\therefore \: p2 - p1 \\ = 0 - 100 \\ = - 100 \: kg \: m {s}^{ - 1}$ So final answer : $\boxed{ \huge{ \red{ - 100 \: kg \: m {s}^{ - 1}}}}$

How much momentum will an object of mass 10 kg transfer to the floor if it falls from a height of 5 m? (g=10m/s)

Answer»

MASS = 10 kg

Height = 5 metres

Momentum transferred to the ground.

The final momentum of the object (just before hitting the ground) will be transferred to the ground, if the collision is considered elastic.

Since the object is dropped, initial velocity of object is zero.

${v}^{2} = {u}^{2} + 2gh \\ = > {v}^{2} = {0}^{2} + (2 \times 10 \times 5) \\ = > {v}^{2} = 100 \\ = > v = \sqrt{100 } \\ = > v = 10 \: m {s}^{ - 1}$

So, final momentum of object just before hitting the ground .

$\therefore \: p1 = m \times v \\ = > p1 = 10 \times 10 \\ = > p1 = 100 \: kg \: m {s}^{ - 1}$

Momentum after hitting the ground :

$p2 = 0 \: kg \: m {s}^{ - 1}$

HENCE, momentum CHANGE of the object is correspondingly transferred to the ground.

So , momentum sent to ground

$\therefore \: p2 - p1 \\ = 0 - 100 \\ = - 100 \: kg \: m {s}^{ - 1}$

So final answer :

$\boxed{ \huge{ \red{ - 100 \: kg \: m {s}^{ - 1}}}}$