Work Power And Energy Ques 3
- A bullet of mass $M$ is fired with a velocity $50 \mathrm{~m} / \mathrm{s}$ at an angle $\theta$ with the horizontal. At the highest point of its trajectory, it collides head-on with a bob of mass $3 M$ suspended by a massless string of length $10 / 3 \mathrm{~m}$ and gets embedded in the bob. After the collision, the string moves through an angle of $120^{\circ}$. Find
$(1988,6 M)$
(a) the angle $\theta$,
(b) the vertical and horizontal coordinates of the initial position of the bob with respect to the point of firing of the bullet. (Take $g=10 \mathrm{~m} / \mathrm{s}^2$ )
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Answer:
Correct Answer: 3.(a) $\theta=30^{\circ}$ (b) The desired coordinates are ( $108.25 \mathrm{~m}, 31.25 \mathrm{~m}$ )
Solution:
Formula:
Coefficient Of Restitution (e):
- (a) At the highest point, velocity of bullet is $50 \cos \theta$. So, by conservation of linear momentum
$ \begin{aligned} & M(50 \cos \theta)=4 M v \\ & \therefore \quad v=\left(\frac{50}{4}\right) \cos \theta \\ & \end{aligned} $
At point $B, T=0$ but $v \neq 0$
Hence, $4 M g \cos 60^{\circ}=\frac{(4 M) v^2}{l}$
or
$ \begin{array}{r} v^2=\frac{g}{2} l=\frac{50}{3} \\ \quad\left(\text { as } l=\frac{10}{3} \mathrm{~m} \text { and } g=10 \mathrm{~m} / \mathrm{s}^2\right) \end{array} $
Also, or
$ \begin{aligned} v^2 & =u^2-2 g h=u^2-2 g\left(\frac{3}{2} l\right) \\ & =u^2-3(10)\left(\frac{10}{3}\right) \end{aligned} $
$ v^2=u^2-100 $
Solving Eqs. (i), (ii) and (iii), we get
$ \cos \theta=0.86 \text { or } \theta=30^{\circ} $
(b) $x=\frac{\text { Range }}{2}=\frac{1}{2}\left(\frac{u^2 \sin 2 \theta}{g}\right)$
$ =\frac{50 \times 50 \times \sqrt{3}}{2 \times 10 \times 2}=108.25 \mathrm{~m} $
$ \begin{aligned} y & =H=\frac{u^2 \sin ^2 \theta}{2 g} \\ & =\frac{50 \times 50 \times 1}{2 \times 10 \times 4}=31.25 \mathrm{~m} \end{aligned} $
Hence, the desired coordinates are $(108.25 \mathrm{~m}, 31.25 \mathrm{~m})$.
BETA
