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Work, Energy, and Power — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsWork, Energy, and Power5 Marks
Question
  1. Discuss the motion of a body in a vertical circle. Find the expressions for the minimum velocity at the lowest and highest points while looping a loop.
  2. A bullet of mass $0.01\ kg$ travelling at a speed of $500m/ s$ strikes a block of mass $2\ kg$ which is suspended by a string of length $5m$. The centre of gravity of the block is found to rise a vertical distance of $0.1 m$. What is the speed of the bullet after it emerges from the block? $(g = 9.8ms^{-2})$
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Answer

Consider the mass $m$ attached to a string of length $l.$ Let the lower most and top most points be marked $A$ and $B$ respectively. Consider a point $P$ where the length $l$ has turned by $\theta$ from the vertical line through $A.$

Centripetal force is provided by the tension and $mg \cos \theta$ acting in opposite directions.
From the diagram, $\text{OA = OP = l,}$
$\text{OT} =\text{l}\cos\theta,\text{AT}=\text{l}(1-\cos\theta)$
  1. If $v_A$ and $v_p$ are velocities at $A$ and $P$ respectively, using
$v^2 = u^2 + 2$ as, we have
$\text{v}^2_\text{p}=\text{v}_\text{A}^2-\text{2g}(\text{AT})$
$\text{v}_\text{}p^2=\text{v}_\text{A}^2-2\text{gl}(1-\cos\theta)$
Velocity at any point $p,$
$\text{v}_\text{p}=\sqrt{\text{v}^2_\text{A}-2\text{gl}(1-\cos\theta)}$
  • At $P, T_p\ -\text{mg}\cos\theta=\frac{\text{mv}^2\text{p}}{\text{l}}$,
$\text{T}_\text{p}=\text{mg}\cos\theta+\frac{\text{mv}^2\text{p}}{\text{l}}$
$\therefore\text{T}_\text{p}=\text{mg}\cos\theta+\frac{\text{m}}{\text{l}}\sqrt{\text{v}_\text{A}^2-2\text{gl}}(\text{l}-\cos\theta)$
By knowing $V_A$ and $\theta$ , one can find the velocity and tension at any point.
  • To perform verticle circle, the tension should be non$-$zero till the mass reaches the top most point $B$
$\therefore\text{T}_\text{B}\geq0\text{ at }\theta=80^\circ$
$\therefore\text{T}_\text{B}=0 $ will be minimum.
Substituting,
$\text{TP}=0\text{ at } \theta =\pi,$ we have
$0 = \text{mg}\cos\pi+ \frac{\text{M}}{\text{l}}[\text{v}_\text{A}^2-2\text{gl}(1-\cos\pi)]$
$0 = -\text{Mg}+\frac{\text{M}}{\text{l}}[\text{v}_\text{A}^2-2\text{gl}(2)]$
$\text{v}_\text{A}^2-4\text{gl}=\text{gl},\text{v}_\text{A}^2=5\text{gl},=\sqrt{5\text{gl}}$
So, the minimum velocity required at lowermost point to perform verticle circle is $\sqrt{5\text{gl}}$.
  • Given: Mass of a bullet $(m) = 0.01\ kg$
Speed of bullet $(v) = 500\ m/ s$
Mass of block $(M) = 2\ kg$
Length of string $(l) = 5\ m$
Vertical height $(h) = 0.1\ m$

Let $V$ be the velocity acquired by block
$\frac{1}{2}\text{MV}=\text{Mgh}$
$\Rightarrow\text{V}=\sqrt{2\text{gh}}$
$=\sqrt{2\times9.8\times0.1}$
​​​​​​​$=104\text{m/s}$
If $V'$ is the speed of bullet on emerging out of the block then by the law of conservation of energy.
$\text{mv}+\text{M}\times0=\text{MV}+\text{mV}'$
$\therefore\text{V}'=\frac{\text{mv}-\text{MV}}{\text{m}}$
$=\frac{0.01\times500-2\times1.4}{0.01}$
$\text{V}'=\frac{5-2.8}{0.01}=\frac{2.2}{0.01}=220\text{ m/s}$

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