Physics Newton
Your question ⤵⤵
Bullots of 40 g mass cach hit a plate at the rate of 100 bullots per second, with a velo

Physics Newton
Your question ⤵⤵
Bullots of 40 g mass cach hit a plate at the rate of 100 bullots per second, with a velocity of 60 m/s and reflect back with a velocity of 30 m/s. The average force acting on
the plate is

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2 thoughts on “Physics Newton<br />Your question ⤵⤵<br />Bullots of 40 g mass cach hit a plate at the rate of 100 bullots per second, with a velo”

  1. [tex] \: \: \: \: \: \: \: \: \: \: \: \: ❍ \large \: \: \underline{\boxed{ \frak{\pmb{ \: \ Question : \: }}}}[/tex]

    Bullots of 40 g mass cach hit a plate at the rate of 100 bullots per second, with a velocity of 60 m/s and reflect back with a velocity of 30 m/s. The average force acting on the plate is ?

    [tex] \: \: \: \: \: \: \: \: \: \: \: \: ❍ \large \: \: \underline{\boxed{ \frak{\pmb{ \: \ given \: : \: }}}}[/tex]

    [tex] \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \purple{: \implies} \rm \: mass = 40g \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \purple{: \implies }\rm \: n = 100 \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \purple{: \implies}\rm \: v_{b}= 60m/s \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \purple{: \implies}\rm \: {v}^{1 _{b}} = 30m/s[/tex]

    [tex] \: \: \: \: \: \: \: \: \: \: \: \: ❍ \large \: \: \underline{\boxed{ \frak{\pmb{ \: \ to \: find \: : \: }}}}[/tex]

    • The average force acting on the plate is ?

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    [tex] \: \: \: \: \: \: \: \: \: \: \: \: ❍ \large \: \: \underline{\boxed{ \frak{\pmb{ \: \ solution \: : \: }}}}[/tex]

    [tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \pink \rightarrow \rm \: pi = nm_{B} v _{B} \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \pink \rightarrow \rm \: n \times 40 \times 60 \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \pink\rightarrow \rm \: 100 \times 40 \times 60 \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \pink\rightarrow\rm240000 \: kg \: m/s[/tex]

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    [tex]\: \: \: \: \: \: \: \: \: \: \: \: ❍ \large \: \: \underline{\boxed{ \frak{\pmb{ \: \ now \: : \: }}}}[/tex]

    [tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red \longmapsto \rm \: pf = nm_{B} {v}^{1} _{B} \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red \longmapsto100 \times 40 \times 30 \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red \longmapsto \rm120000 \: Kg \: m/s[/tex]

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    [tex]\: \: \: \: \: \: \: \: \: \: \: \: ❍ \large \: \: \underline{\boxed{ \frak{\pmb{ \: \ then\: : \: }}}}[/tex]

    [tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \green⇁ \rm \: p = pi – pf \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \green⇁ \rm \: 120000 \: Kg \: m/s[/tex]

    So, Finally :

    [tex] \rm \: Average \: force = \dfrac{change \: in \: momentum}{time \: taken} \\ \\ \\ \rm \implies \dfrac{120000}{1} \\ \\ \\ \rm\implies120000 \: N[/tex]

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    Reply
  2. Answer :-

    Average force acting on the plate is a retarding force of 360 N .

    Explanation :-

    We have :-

    → Mass of each bullet = 40g = 0.04 kg

    → Number of bullets = 100

    → Time (t) = 1 second

    → Initial velocity (u) = 60 m/s

    → Final velocity (v) = 30 m/s

    [As the bullets reflect back so ‘v’ is -ve].

    _________________________________

    Mass of 100 bullets (m) :-

    = 100(0.04)

    = 4 kg

    According to Newton’s 2nd law of motion, we know that :-

    F = (mv mu)/t

    Substituting values, we get :-

    ⇒ F = [4(-30) – 4(60)]/1

    ⇒ F = [-120 – 240]/1

    ⇒ F = -360/1

    F = -360 N

    [Here, -ve sign represents retarding force ]

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