Question

Solve Plz...zZzzzzz​

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

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

Now we shall find the average force!!

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

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

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

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

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

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

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

Answer 2

Answer:

GIVEN :-

Mass = 250 g

Speed = 2 m/s

Time of contact = 10⁻³ sec

The average force exerted by the wall on the ball = ?

━━━━━━━━━━━━━━

\begin{gathered}\begin{gathered}\begin{gathered}:\implies \sf{ \Delta p = mv + mv}\\\\\end{gathered} \end{gathered}\end{gathered}:⟹Δp=mv+mv

\begin{gathered}\begin{gathered}\begin{gathered}:\implies \sf{ \Delta p = 2(mv)}\\\\\end{gathered} \end{gathered}\end{gathered}:⟹Δp=2(mv)

Now we shall find the average force!!

\begin{gathered}\begin{gathered}\begin{gathered}:\implies \sf{F = \dfrac{dp}{dt} }\\\\\end{gathered} \end{gathered}\end{gathered}:⟹F=dtdp

\begin{gathered}\begin{gathered}\begin{gathered}:\implies \sf{F = \dfrac{\Delta p}{\Delta t} }\\\\\end{gathered} \end{gathered}\end{gathered}:⟹F=ΔtΔp

\begin{gathered}\begin{gathered}\begin{gathered}:\implies \sf{F = \dfrac{2mv}{\Delta t} }\\\\\end{gathered} \end{gathered}\end{gathered}:⟹F=Δt2mv

\begin{gathered}\begin{gathered}\begin{gathered}:\implies \sf{F = \dfrac{2 \times 0.25 \times 2}{ {10}^{ - 3}} }\\\\\end{gathered} \end{gathered}\end{gathered}:⟹F=10−32×0.25×2

\begin{gathered}\begin{gathered}:\implies \sf F = \dfrac{2 \times 250 \times 10^{-3} \times 2}{ {10}^{ - 3}} \\\\\end{gathered}\end{gathered}:⟹F=10−32×250×10−3×2

\begin{gathered}\begin{gathered}\begin{gathered}:\implies \sf F = \dfrac{1000\times \cancel{10^{-3}}}{\cancel{{10}^{ - 3}}} \\\\\end{gathered} \end{gathered}\end{gathered}:⟹F=10−31000×10−3

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