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JEE Advanced 2024Paper 2
Exam
JEE Advanced
Year
2024
Paper
Paper 2
Subject
Physics
Chapter
Motion in One Dimension

JEE Advanced 2024 · Paper 2 · Physics · Motion in One Dimension

A ball is thrown from the location (x_0, y_0)=(0,0) of a horizontal...

A ball is thrown from the location (x0,y0)=(0,0)\left(x_0, y_0\right)=(0,0) of a horizontal playground with an initial speed v0v_0 at an angle θ0\theta_0 from the +x+x-direction. The ball is to be hit by a stone, which is thrown at the same time from the location (x1,y1)=(L,0)\left(x_1, y_1\right)=(L, 0). The stone is thrown at an angle \left(180-\theta_1\right)fromthefrom the+x−directionwithasuitableinitialspeed.Forafixed-direction with a suitable initial speed. For a fixedv_0,when(θ0,θ1)=(45∘,45∘),, when \left(\theta_0, \theta_1\right)=\left(45^{\circ}, 45^{\circ}\right), the stone hits the ball after timeT_1, and when(θ0,θ1)=(60∘,30∘)when \left(\theta_0, \theta_1\right)=\left(60^{\circ}, 30^{\circ}\right), it hits the ball after time T2T_2. In such a case, \left(T_1 / T_2\right)^2 is _______ .A ball is thrown from the location (x_0, y_0)=(0,0) of a horizontal playground with an initial speed v_0 at an angleA ball is thrown from the location (x_0, y_0)=(0,0) of a horizontal playground with an initial speed v_0 at an angle

Correct answer

0

Step-by-step solution

For case I : A ball is thrown from the location (x_0, y_0)=(0,0) of a horizontal playground with an initial speed v_0 at an angle arel =0a_{\text{rel }}=0 For collision v02=v2∴\frac{\mathrm{v}_0}{\sqrt{2}}=\frac{\mathrm{v}}{\sqrt{2}}\therefore\mathrm{v}=\mathrm{v}0_0 So T1=Lv02+v2∴τ1=L2v0…..(1)\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 ball is thrown from the location (x_0, y_0)=(0,0) of a horizontal playground with an initial speed v_0 at an angle arel =0a_{\text{rel }}=0 For collision, v032=v2\frac{\mathrm{v}_0 \sqrt{3}}{2}=\frac{\mathrm{v}}{2}∴v=3v0 So, T2=Lv02+v32 T2=Lv02+3v02∴T2=L2v0……..(2) so, (T1 T2)2=(2)2=2⇒(T1 T2)2=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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