Question

A mass m falls freely from rest.The linear momentum after it has fallen through a height h is

Answer 1

Answer:

Given:

A mass m freely falls through a height h.

To find:

Linear momentum after the object has fallen by height h.

Calculation :

We will be using basic equations of Kinematics .

v² = u² + 2gh

=> v² = 0² + 2 × g × h

=> v² = 2gh

=> v √(2gh).

Linear momentum is given by the product of mass and velocity at a particular point.

Momentum = mass × vel.

=> Momentum = [m × √(2gh) ].

Answer 2

${ \huge \bf { \mid{ \overline{ \underline{Question}}} \mid}}$

A mass m falls freely from rest.The linear momentum after it has fallen through a height h is?

$\huge{\bold{\underline{\underline{Answer}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Initial velocity (u) = 0 m/s.
• Height it Falls = h
• Mass of the body = m.

$\huge{\bold{\underline{Explanation:-}}}$

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From Third kinematic equation,

$\large{\boxed{\tt v^2 - u^2 = 2as}}$

• Here a = g (Free fall)
• s = h

Substituting,

$\large{\tt \leadsto v^2 - (0)^2 = 2gh}$

$\large{\tt \leadsto v^2 = 2gh}$

$\large{\tt \leadsto v = \sqrt{2gh} \: ----(1)}$

$\large{\boxed{\tt v = \sqrt{2gh}}}$

Therefore, velocity of the body is v = √2gh.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From Change in Momentum Formula,

$\large{\boxed{\tt P = mv - mu}}$

Simplifying,

$\large{\tt \leadsto P = m(v - u)}$

Substituting the values of velocity,

$\large{\tt \leadsto P = m \times ( \sqrt{2gh} - 0)}$

$\large{\tt \leadsto P = m \times \sqrt{2gh}}$

$\huge{\boxed{\boxed{\tt P = m.\sqrt{2gh}}}}$

So, the Momentum of the body will be P = m × √2gh.

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