Physics, asked by amritakarthikeyan8, 8 months ago

A ball of mass 2 kg is dropped from a height of 5 m. On striking the ground it rebounds with same speed

with which it struck the ground. Find the magnitude of net change in its momentum just after striking the

ground. [Take, g = 10 m/s2

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.pls give me the correct ans pls....​

Answers

Answered by ExᴏᴛɪᴄExᴘʟᴏʀᴇƦ
22

\displaystyle\large\underline{\sf\red{Given}}

✭ Mass of a ball is 2 kg

✭ It is dropped from a height of 5 m

✭ Initial Velocity of the ball is 0 m/s

✭ Gravitational Acceleration = 10 m/s²

\displaystyle\large\underline{\sf\blue{To \ Find}}

◈ The magnitude of net change in its momentum after it strikes the ground?

\displaystyle\large\underline{\sf\gray{Solution}}

Do here we shall first find the velocity of the body with the help of the third equation of motion that is,

\displaystyle\sf \underline{\boxed{\sf v^2-u^2=2as}}

And then we shall find the change in Momentum with the help of,

\displaystyle\sf \underline{\boxed{\sf Change \ in \ momentum = mv-mu}}

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\underline{\bigstar\:\textsf{According to the given Question :}}

So now the velocity with which the ball strikes the ground will be,

\displaystyle\sf v^2-u^2=2as\\

\displaystyle\sf v^2-0^2= 2\times10\times 5\\

\displaystyle\sf v^2=100\\

\displaystyle\sf v =\sqrt{100}\\

\displaystyle\sf \orange{v = 10 \ m/s}

✪ The ball here is falling down which means it's initial velocity (u) should be Negative and in this case it is -10 m/s [u = -10 m/s]

✪ Also Given that the ball rebounds with the same velocity but when it rebounds it goes up which means it's final Velocity (v) will be positive and in this case it is 10 m/s [v = 10 m/s]

Change in Momentum is given by,

\displaystyle\sf Change \ in \ Momentum = mv-mu\\

\displaystyle\sf 2\times 10-2\times(-10)\\

\displaystyle\sf 20-(-20)\\

\displaystyle\sf 20+20\\

\displaystyle\sf \pink{Change \ in \ Momentum = 40 \ kgm/s}\\

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Answered by sagarbricks0024
0

change in momentum

40 kgm/s

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