Question

a ball of mass 250gm moving with a speed of 2m/s hits a ball normally .it rebounds with same speed if the two were in the contact for 10^-3 seconds.The average force exerted by the wall on the ball is?.PLEASE GIVE PROPER ANSWERS​
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Answer 1

Given

• Mass = 250 g
• Speed = 2 m/s
• Time of contact = 10⁻³ sec
• The average force exerted by the wall on the ball = ?

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$:\implies \sf{ \Delta p = mv + mv}$

$:\implies \sf{ \Delta p = 2(mv)}$

• Now we shall find the average force!!

$:\implies \sf{F = \dfrac{dp}{dt} }$

$:\implies \sf{F = \dfrac{\Delta p}{\Delta t} }$

$:\implies \sf{F = \dfrac{2mv}{\Delta t} }$

$:\implies \sf{F = \dfrac{2 \times 0.25 \times 2}{ {10}^{ - 3}} }$

$:\implies \sf F = \dfrac{2 \times 250 \times 10^{-3} \times 2}{ {10}^{ - 3}}$

$:\implies \sf F = \dfrac{1000\times \cancel{10^{-3}}}{\cancel{{10}^{ - 3}}}$

$:\implies \underline{\boxed{\mathfrak{\pink{ f = 1000 \: n}}}}$

Answer 2

Aɴsᴡᴇʀ :-

• Force exerted by the wall = $\boxed{\textsf{\textbf{\red{1000 N}}}}$

Gɪᴠᴇɴ :-

• Mass of the ball (m) = 250g = 0.25kg

• Speed of the ball (u) = 2 m/s

• Speed when rebounce (v) = -2 m/s

• Time in contact (∆t) = $\sf{10^{-3}}$

Tᴏ Fɪɴᴅ :-

• Average force exerted by the wall.

Fᴏʀᴍᴜʟᴀ Usᴇᴅ :-

• $\boxed{\sf{F = \dfrac{∆p}{∆t}}}$

• $\boxed{\sf{∆p = mv - mu}}$

Sᴏʟᴜᴛɪᴏɴ :-

➪ $\sf{F = \dfrac{∆p}{∆t}}$

➪ $\sf{F = \dfrac{m(v-u)}{∆t}}$

➪ $\sf{F = \dfrac{0.25(-2-2)}{10^{-3}}}$

➪ $\sf{F = -\dfrac{1}{10^{-3}}}$

➪ $\sf{F = -1000 N}$

Here, Negative sign indicates direction (i.e, wall on the ball)

So, $\sf{F = }$ $\textbf{1000 N}$