Question

3. A force of 10 N acts on a body of mass 2 kg for 3 seconds, initially at rest. Calculate:
(i) Velocity acquired by the body
(ii) Change in momentum of the body.​

Answer 1

$\large \rm {\underbrace{\underline{Elucidation:-}}}$

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$\dag \sf \red {\underline{\underline{Provided\: that:}}}$

$\to \tt {Force(f)=10N}$

$\to \tt {Mass\: of\: the\: body(m)=2kg}$

$\to \tt {Time\:interval(t)=3s}$

➻It is said that the body is intially at rest.

➻so,

$\to \tt {Intial\:velocity(u)=0}$

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$\dag \sf \blue {\underline{\underline{To\: determine:}}}$

(I) Velocity acquired by the body,i.e,

$\to \tt {Final\: Velocity (v)=?}$

(II) change in momentum,i.e,

$\to \tt {∆p=?}$

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$\dag \sf \orange {\underline{\underline{Using\: 1^{st}\: equation\: of\: motion}}}$

$\implies \tt {\fbox{\underline{v=u+at}}}$

➻But to use this kinematic equation,we need to calculate acceleration.

➻To calculate acceleration,we know,

$\to \tt {\fbox{Force=mass×acceleration}}$

$\to \tt {F=m×a}$

➻supplanting the given values,

$\to \tt {10N=2kg×a}$

$\large \to \tt {a=\frac{10}{2}}$

$\to \tt \green {\boxed{\underline{Acceleration=5m/s^{2}}}}$

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➻As the acceleration is determined we can calculate final velocity.

$\to \tt {v=u+at}$

$\to \tt {v=0+5×3}$

$\colon \implies \tt \green {\fbox{\underline{v=15m/s}}}$

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$\dag \sf \pink {\underline{\underline{Change\: in\: momentum:}}}$

➻Change in momentum=Final momentum-inital momentum

$\to \tt {∆p=mv-mu}$

$\to \tt {∆p=2kg×15m/s-2kg×0}$

$\to \tt {∆p=30kgm/s-0}$

$\to \tt \green{\fbox{\underline{∆p=30kgm/s}}}$

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$\dag \sf \purple {\underline{\underline{Thereupon,}}}$

➻The Velocity acquired by the body is 15m/s and the change in momentum of the body is 30kgm/s respectively.

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