JEE MAIN - Physics (2022 - 27th June Morning Shift - No. 26)
A pendulum of length 2 m consists of a wooden bob of mass 50 g. A bullet of mass 75 g is fired towards the stationary bob with a speed v. The bullet emerges out of the bob with a speed $${v \over 3}$$ and the bob just completes the vertical circle. The value of v is ___________ ms$$-$$1. (if g = 10 m/s2).
Answer
10
Explanation
Let minimum velocity at the lowest point for the bob to compleate the circular path = v'
And we know, v' = $\sqrt{5 r g}$
From the conservation of momentum, we have
$$ 75 \times 10^{-3} \times v=50 \times 10^{-3} \times v^{\prime}+75 \times 10^{-3} \times \frac{v}{3} $$
$75 \times 10^{-3} v=50 \times 10^{-3} \sqrt{5 r g} \times\left(75 \times 10^{-3} \times \frac{v}{3}\right)$
According to question, the bob completes a vertical circle of $2 \mathrm{~m}$ radius, therefore
$$ \begin{aligned} & r=2 \mathrm{~m}, g=10 \mathrm{~ms}^{-2} \\\\ & 75 \times 10^{-3} v=50 \times 10^{-3} \times \sqrt{5 \times 2 \times 10}+75 \times 10^{-3} \times \frac{v}{3} \\\\ & 75 \times 10^{-3}\left(\frac{2 v}{3}\right)=50 \times 10^{-3} \times 10 \\\\ & 150 \times 10^{-3} \times v=(150) \times 10^{-2} \\\\ & v=10 \mathrm{~m} / \mathrm{s} \end{aligned} $$
Note :
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At the point $3$, both the tension $T_{3}$ and the weight $m g$ of the body act towards the centre of the circle. So $T_{3}+m g$ provides the centripetal force necessary for the rotation of the body.
$$ \therefore T_{3}+m g=\frac{m v_{3}^{2}}{r} $$
At the point $1$, the tension $T_{1}$ acts vertically upwards i.e., towards the centre of the circle and the weight $m g$ of the body acts vertically downwards i.e., in an opposite direction. So $T_{1}-m g$ provides the necessary centripetal force here.
$$ \therefore T_{1}-m g=\frac{m v_{1}{ }^{2}}{r} $$
If the tension in the string just vanishes at $3$ i.e., if $T_{3}=0$, then
$$ m g=\frac{m v_{3}^{2}}{r} \text { or, } v_{3}=\sqrt{g r} $$
If the velocity of the body at the highest point $3$ be less than $\sqrt{g r}$, the string will slack and the body will drop down instead of rotating in the circular path. So this minimum velocity of the body at the highest point is called the critical velocity.
Minimum velocity at the lowest point for maintaining the critical velocity :
Now, as the body goes from $3$ to $1$, its height increases by $2 r$. So its potential energy increases by $m g \times 2 r$. From the principle of conservation of mechanical energy, we have
The K.E. of the body at $3$ - Its K.E. at $1=$ Increase in P.E.
or, $1 / 2 m v_{3}^{2}-1 / 2 m v_{1}^{2}=2 m g r$
or, $v_{3}^{2}=v_{1}^{2}+4 g r$
When $v_{3}=\sqrt{g r}, v_{1}$ is minimurn
$\therefore\left(v_{1}\right)_{\min }=\sqrt{g r+4 g r}=\sqrt{5 g r}$
Minimum tension :
When the body moves with the critical velocity at the highest point, the tension in the string becomes zero; then $m g=\frac{m v_{3}^{2}}{r}$ When this condition is satisfied, the tension in the string at the lowest point $1$ becomes minimum.
So, $$ \left(T_{1}\right)_{\min }=\frac{m\left(v_{1}\right)^{2} \min }{r}+\frac{m\left(v_{3}\right)^{2} \min }{r}=\frac{m}{r}(5 g r+g r)=6 \mathrm{mg} $$
And we know, v' = $\sqrt{5 r g}$
From the conservation of momentum, we have
$$ 75 \times 10^{-3} \times v=50 \times 10^{-3} \times v^{\prime}+75 \times 10^{-3} \times \frac{v}{3} $$
$75 \times 10^{-3} v=50 \times 10^{-3} \sqrt{5 r g} \times\left(75 \times 10^{-3} \times \frac{v}{3}\right)$
According to question, the bob completes a vertical circle of $2 \mathrm{~m}$ radius, therefore
$$ \begin{aligned} & r=2 \mathrm{~m}, g=10 \mathrm{~ms}^{-2} \\\\ & 75 \times 10^{-3} v=50 \times 10^{-3} \times \sqrt{5 \times 2 \times 10}+75 \times 10^{-3} \times \frac{v}{3} \\\\ & 75 \times 10^{-3}\left(\frac{2 v}{3}\right)=50 \times 10^{-3} \times 10 \\\\ & 150 \times 10^{-3} \times v=(150) \times 10^{-2} \\\\ & v=10 \mathrm{~m} / \mathrm{s} \end{aligned} $$
Note :
At the point $3$, both the tension $T_{3}$ and the weight $m g$ of the body act towards the centre of the circle. So $T_{3}+m g$ provides the centripetal force necessary for the rotation of the body.
$$ \therefore T_{3}+m g=\frac{m v_{3}^{2}}{r} $$
At the point $1$, the tension $T_{1}$ acts vertically upwards i.e., towards the centre of the circle and the weight $m g$ of the body acts vertically downwards i.e., in an opposite direction. So $T_{1}-m g$ provides the necessary centripetal force here.
$$ \therefore T_{1}-m g=\frac{m v_{1}{ }^{2}}{r} $$
If the tension in the string just vanishes at $3$ i.e., if $T_{3}=0$, then
$$ m g=\frac{m v_{3}^{2}}{r} \text { or, } v_{3}=\sqrt{g r} $$
If the velocity of the body at the highest point $3$ be less than $\sqrt{g r}$, the string will slack and the body will drop down instead of rotating in the circular path. So this minimum velocity of the body at the highest point is called the critical velocity.
Minimum velocity at the lowest point for maintaining the critical velocity :
Now, as the body goes from $3$ to $1$, its height increases by $2 r$. So its potential energy increases by $m g \times 2 r$. From the principle of conservation of mechanical energy, we have
The K.E. of the body at $3$ - Its K.E. at $1=$ Increase in P.E.
or, $1 / 2 m v_{3}^{2}-1 / 2 m v_{1}^{2}=2 m g r$
or, $v_{3}^{2}=v_{1}^{2}+4 g r$
When $v_{3}=\sqrt{g r}, v_{1}$ is minimurn
$\therefore\left(v_{1}\right)_{\min }=\sqrt{g r+4 g r}=\sqrt{5 g r}$
Minimum tension :
When the body moves with the critical velocity at the highest point, the tension in the string becomes zero; then $m g=\frac{m v_{3}^{2}}{r}$ When this condition is satisfied, the tension in the string at the lowest point $1$ becomes minimum.
So, $$ \left(T_{1}\right)_{\min }=\frac{m\left(v_{1}\right)^{2} \min }{r}+\frac{m\left(v_{3}\right)^{2} \min }{r}=\frac{m}{r}(5 g r+g r)=6 \mathrm{mg} $$
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