A mass m falls freely from rest. The linear momentum after it has fallen through a height h is: (g = acceleration due to gravity)

Question

A mass m falls freely from rest. The linear momentum after it has fallen through a height h is: (g = acceleration due to gravity) (A)  √mgh  (B)  m√2gh  (C)  m√gh  (D) zero

Tardigrade Admin · Accepted Answer

When the body is falling freely from height h, then from third equation of motion v2-u2=2gh (∵ u=0) ∴ v2=2gh⇒ v=√2gh Now, linear momentum of body is given p=m text v From Eqs.(i) and (ii) Momentum =m√2gh

Q. A mass m falls freely from rest. The linear momentum after it has fallen through a height h is: (g = acceleration due to gravity)

$m g h$

$m 2 g h$

$m g h$

zero

When the body is falling freely from height h, then from third equation of motion $v^{2} - u^{2} = 2 g h$ $(∵ u = 0)$ $∴$ $v^{2} = 2 g h \Rightarrow v = 2 g h$ Now, linear momentum of body is given $p = m v$ From Eqs.(i) and (ii) Momentum $= m 2 g h$

Q. A mass m falls freely from rest. The linear momentum after it has fallen through a height h is: (g = acceleration due to gravity)

mgh​

m2gh​

mgh​

zero

When the body is falling freely from height h, then from third equation of motion v2−u2=2gh (∵u=0) ∴ v2=2gh⇒v=2gh​ Now, linear momentum of body is given p=mv From Eqs.(i) and (ii) Momentum =m2gh​

$m g h$

$m 2 g h$

$m g h$

When the body is falling freely from height h, then from third equation of motion $v^{2} - u^{2} = 2 g h$ $(∵ u = 0)$ $∴$ $v^{2} = 2 g h \Rightarrow v = 2 g h$ Now, linear momentum of body is given $p = m v$ From Eqs.(i) and (ii) Momentum $= m 2 g h$