Question

A bullet of mass 0.5kg and moving with a velocity 100m/s is stopped with in 0.1m of target. Find the average resistance force offered by the target.

Answer 1

Given data : bullet of mass 0.5 kg and moving with a velocity 100 m/s is stopped with in 0.1 m of target.

To find : The average resistance force offered by the target ?

Solution : according to given data,

• Mass of bullet ( m ) = 0.5 kg
• Initial velocity of bullet ( u ) = 100m/s
• Final velocity of the bullet ( v ) = 0 {Because the bullet stopped}
• Displacement of the bullet ( s ) = 0.1 m

Now, we need to find acceleration ( a ) of the bullet, by using kinematical equation;

⟹ v² = u² + 2 * a * s

⟹ 0 = ( 100 )² + 2 * a * 0.1

⟹ 0 = 10000 + 0.2 * a

⟹ 0 - 10000 = 0.2 * a

⟹ - 10000 = 0.2 * a

⟹ a = - 10000/0.2

⟹ a = - 50000

Now, for average resistance force offered by the target,

⟹ Force = mass * acceleration

⟹ Force = 0.5 * ( - 50000 )

⟹ Force = - 25000 N

Answer : The average resistance force offered by the target is - 25000 N.

More info : The negative sign indicates that the forces are in opposite directions.

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Answer 2

Given :-

• Mass of bullet,m = 0.5 kg
• Initial velocity of bullet, u= 100 m/s
• Final velocity of the bullet,v = 0
• Displacement of the bullet, s = 0.1 m

To find :-

• The average resistance force offered by the target

Solution :

• Firstly, we have to find acceleration of the bullet to the average resistance force offered by the target.

$\small {\mathfrak {\underline { Using\:3rd \: equation \: of \: motion \: :- }}}$

$\: \: \: \: \: \: \: : \implies \underline{ \boxed { \pink{{\frak{v² = u² + 2as}}}}}$

$\small\underline{\pmb{\sf \:Substituting \: Values :-}}$

$\: \: \: \: \: \: \: : \implies \sf \: 0= ( 100 )² + 2 \times a \times 0.1$

$\: \: \: \: \: \: \: : \implies \sf \: 0= ( 100 )² + 0.2 \times a$

$\: \: \: \: \: \: \: : \implies \sf \: a \times 0.2= 10000$

$\: \: \: \: \: \: \: : \implies \sf \: a \times 0.2= -10000$

$\: \: \: \: \: \: \: : \implies \sf \: a =\cancel{\dfrac{ -10000}{0.2}}$

$\: \: \: \: \: \: \: : \implies \underline{ \boxed { \pink{{\frak{a = -50000 m/s²}}}}}$

(Since, Negative acceleration is retardation )

$\therefore\:\underline{\textsf{Acceleration is \textbf{-50000 \:m/s²}}}.$

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀For average resistance force:-

$\: \: \: \: \: \: \: : \implies \underline{ \boxed { \purple{{\frak{F =ma }}}}}$

$\sf \: \: \: \: \: \: \: :\implies Force = 0.5 \times - 50000$

$\: \: \: \: \: \: \: : \implies \underline{ \boxed { \purple{{\frak{F =-25000N }}}}}$

(Here,negative sign indicates force applied on the opposite direction.)

$\therefore\:\underline{\textsf{ Average resistance force is \textbf{-25000N}}}.$