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JEE Advanced · Physics · 3. Motion in 1D

A ball is thrown from the location \(\left(x_0, y_0\right)=(0,0)\) of a horizontal playground with an initial speed \(v_0\) at an angle \(\theta_0\) from the \(+x\)-direction. The ball is to be hit by a stone, which is thrown at the same time from the location \(\left(x_1, y_1\right)=(L, 0)\). The stone is thrown at an angle \(\left(180-\theta_1\right)\) from the \(+x\)-direction with a suitable initial speed. For a fixed \(v_0\), when \(\left(\theta_0, \theta_1\right)=\left(45^{\circ}, 45^{\circ}\right)\), the stone hits the ball after time \(T_1\), and when \(\left(\theta_0, \theta_1\right)=\left(60^{\circ}, 30^{\circ}\right)\), it hits the ball after time \(T_2\). In such a case, \(\left(T_1 / T_2\right)^2\) is _______ .

  1. A 2
  2. B 4
  3. C 6
  4. D 8
Verified Solution

Answer & Solution

Correct Answer

(A) 2

Step-by-step Solution

Detailed explanation

For case I :

\(a_{\text {rel }}=0\)
For collision \(\frac{\mathrm{v}_0}{\sqrt{2}}=\frac{\mathrm{v}}{\sqrt{2}}\)
\(\therefore \mathrm{v}=\mathrm{v}_0\)
So \(\mathrm{T}_1=\frac{\mathrm{L}}{\frac{\mathrm{v}_0}{\sqrt{2}}+\frac{\mathrm{v}}{\sqrt{2}}}\)
\(\therefore \tau_1=\frac{\mathrm{L}}{\sqrt{2} \mathrm{v}_0} \ldots . .(1)\)
For case II,

\(a_{\text {rel }}=0\)
For collision, \(\frac{\mathrm{v}_0 \sqrt{3}}{2}=\frac{\mathrm{v}}{2}\)
\(\begin{aligned}
& \therefore \mathrm{v}=\sqrt{3} \mathrm{v}_0 \\
& \text { So, } \mathrm{T}_2=\frac{\mathrm{L}}{\frac{\mathrm{v}_0}{2}+\mathrm{v} \frac{\sqrt{3}}{2}} \\
& \mathrm{~T}_2=\frac{\mathrm{L}}{\frac{\mathrm{v}_0}{2}+\frac{3 \mathrm{v}_0}{2}} \\
& \therefore \mathrm{T}_2=\frac{\mathrm{L}}{2 \mathrm{v}_0} \ldots \ldots . .(2) \\
& \text { so, }\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)^2=(\sqrt{2})^2=2 \Rightarrow\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)^2=2
\end{aligned}\)
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