Kinematics Ques 7
- On a frictionless horizontal surface, assumed to be the $x-y$ plane, a small trolley $A$ is moving along a straight line parallel to the $y$-axis (see figure) with a constant velocity of $(\sqrt{3}-1) \mathrm{m} / \mathrm{s}$. At a particular instant when the line $O A$ makes an angle of $45^{\circ}$ with the $x$-axis, a ball is thrown along the surface from the origin $O$. Its velocity makes an angle $\phi$ with the $x$-axis and it hits the trolley.
$(2002,5 M)$
(a) $6 \mathrm{~m}$
(b) $3 \mathrm{~m}$
(c) $10 \mathrm{~m}$
(d) $9 \mathrm{~m}$
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Answer:
Correct Answer: 7.(a) $45^{\circ}$ (b) $2 \mathrm{~m} / \mathrm{s}$
Solution:
- (a) Let $A$ stands for trolley and $B$ for ball. Relative velocity of $B$ with respect to $A\left(\mathbf{v} _ {B A}\right)$ should be along $O A$ for the ball to hit the trolley. Hence, $\mathbf{v} _ {B A}$ will make an angle of $45^{\circ}$ with positive $X$-axis.
$ \theta=45^{\circ} $
(b) Let $v=$ absolute velocity of ball.
$ \begin{aligned} \phi & =\frac{4 \theta}{3}=\frac{4}{3}\left(45^{\circ}\right)=60^{\circ} \rightarrow \text { with } X \text {-axis } \\ \therefore \quad \mathbf{v} _ B & =(v \cos \theta) \hat{\mathbf{i}}+(v \sin \theta) \hat{\mathbf{j}} \\ & =\frac{v}{2} \hat{\mathbf{i}}+\frac{\sqrt{3} v}{2} \hat{\mathbf{j}} \\ \mathbf{v} _ A & =(\sqrt{3}-1) \hat{\mathbf{j}} \\ \therefore \quad \mathbf{v} _ {B A} & =\frac{v}{2} \hat{\mathbf{i}}+\left(\frac{\sqrt{3} v}{2}-\sqrt{3}+1\right) \hat{\mathbf{j}} \end{aligned} $
Since, $\mathbf{v} _ {B A}$ is at $45^{\circ}$
$\therefore \quad \frac{v}{2}=\frac{\sqrt{3} v}{2}-\sqrt{3}+1$
or $\quad v=2 \mathrm{~m} / \mathrm{s}$
BETA
