Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

A ball of mass m falls vertically to the ground from a height $\mathrm{h_{1}}$ and rebounds to a height $\mathrm{h_{2}}$ . The change in momentum of the ball on striking the ground is :-
Option: 1
$\mathrm{m \sqrt{2} g\left(h_1+h_2\right)}$

Option: 2
$\mathrm{n \sqrt{2 g\left(m_1+m_2\right)}}$

Option: 3
$\mathrm{m g\left(h_1-h_2\right)}$

Option: 4
$\mathrm{m\left(\sqrt{2 g h_1}-\sqrt{2 g h_2}\right)}$

Answers (1)

best_answer

Let $V_1$ = Velocity when height of free fall is $\mathrm{h_1}$ ,

$\mathrm{V_2}$ = Velocity when height of free rise is $\mathrm{h_2}$

$\mathrm{\therefore \quad v_1^2=u^2+2 g h_1}$ for free fall

Or $\mathrm{\therefore \quad v_1^2=0+2gh}=2gh$

For free rise after impact on ground

$\mathrm{ 0=V_2^2-2 g h_2 \text { or } V_2^2=2 g h_2}$

$\mathrm{ Initial\: \: momentum =m v, }$

$\mathrm{ Final \: \: momentum =m v_2 }$

Now, change in momentum $\mathrm{ \begin{aligned} & =m\left(V_1-V_2\right) \\ & =m\left(\sqrt{2 g h_1}-\sqrt{2 g h_1}\right) \end{aligned} }$

Posted by

SANGALDEEP SINGH

View full answer

NEET 2024 Most scoring concepts

Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

NEET Paper Leak: Court extends CBI custody of accused Manisha Hawaldar by 2 days

anam-khan

'PM Modi personally supervising the situation': Centre tells SC in NEET case

anam-khan

NEET Paper Leak Row: NTA informs SC of security overhaul, institutional reforms

Latest Question