Question

A rubber ball of mass 50 g falls from height of 1 m and rebound to height of 0.5 m . Find the impulse and average force between the ball and the ground . If the time for which was ball in the contact with ground is 0.1 sec.
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Answer 1

AnswEr :

$\bullet \: \textsf {Impulse is{ \textbf{0.225 Ns}}}$

$\bullet \: \textsf {Force is{ \textbf{2.25 N}}}$

GivEn :

• $\sf Mass \: of \: Ball \: = 50 \: g$

• $\sf \: Height = 1 \: m$

• $\sf Time = 0.1 s$

• $\sf Initial\: velocity = 0 \: m/s$

• $\sf \: Rebound \: height = 0.5 \: m$

To find :

• Impulse

• Average force

SoluTion :

⠀⠀⠀We'll find the final velocity (v) at first !

Now,

Apply the formula of equation of motion,

$: \implies\sf \: \: {v}^{2} = {u}^{2} + 2gh$

$: \implies\sf \: \: {v}^{2} = 0+2 \times 9.8 \times 0.1$

$: \implies\sf \: \: {v}^{2} = \sqrt{1.96 }$

$: \implies\sf \: \: { \boxed{ \boxed{ \sf{ \red{v = 1.4 m/s}}}}} \: \bigstar$

Now, it is also given that in the second case the final velocity of the Ball is 0 m/s & we've to find the initial velocity!

Now,

$:\implies \sf \: \: {v}^{2} = {u}^{2} -2gh$

$:\implies \sf \: \:0= {u}^{2} - 2 \times 9.8 \times 0.5$

$:\implies \sf \: \: {u}^{2} =9.8$

$:\implies \sf \: \:{ \boxed{ \boxed{ \sf{ \purple{u=3.1 \: m/s}}}}} \: \bigstar$

Now, we know the relation b/w impulse and change in the linear momentum!

$\bigstar \sf \: Req \: Impulse = Change \: in \: linear \: momentum$

$: \implies \sf \: \: mv - m( - u)$

$: \implies \sf \: \:m(v+u)$

$: \implies \sf \: \:0.05 \times (1.4+3.1)$

$: \implies \sf \: \:{ \boxed{ \boxed{ \bf{ \pink{ \mid \mid0.225 \: Ns \mid \mid}}}}} \: \bigstar$

Apply formula of force required,

$\: \: \: \: \: \: \star \: \boxed{\sf{ \blue{Force = \dfrac{Impulse}{Time} }}}$

$\sf : \implies \: Force = \dfrac{Impulse}{Time}$

$\sf : \implies \: Force = \dfrac{0.225}{0.1}$

$\sf : \implies \: { \boxed{ \boxed{ \bf{ \red{ \mid\mid \: Force = 2.25 \: N \mid \mid}}}}} \: \bigstar$

⠀⠀Therefore,

$\bullet \: \textsf {Impulse is{ \textbf{0.225 Ns}}}$

$\bullet \: \textsf {Force is{ \textbf{2.25 N}}}$