Verified Solution • 2025–26 Session moderate

QJEEM22Phe954: JEE Main 2022 Physics - Class 11 JEEM22Phe954 Solution: A body of mass is projected at an angle of with the horizontal. …

A body of mass 10 kg10 \mathrm{~kg} is projected at an angle of 45∘45^{\circ} with the horizontal. The trajectory of the body is observed to pass through a point (20,10)(20,10) . If T\mathrm{T} is the time of flight, then its momentum vector, at time t=T2\mathrm{t}=\frac{\mathrm{T}}{\sqrt{2}} , is _____________.

[Take g=10 m/s2\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2} ]

Multiple Choice Options

  • A 100i^+(1002−200)j^100 \hat{i}+(100 \sqrt{2}-200) \hat{j}
  • B 1002i^+(100−2002)j^100 \sqrt{2} \hat{i}+(100-200 \sqrt{2}) \hat{j}
  • C 100i^+(100−2002)j^100 \hat{i}+(100-200 \sqrt{2}) \hat{j}
  • D 1002i^+(1002−200)j^100 \sqrt{2} \hat{i}+(100 \sqrt{2}-200) \hat{j}

Step-by-Step Verified Solution

m = 10 kg

θ\theta = 45 ∘^\circ

y=xtan⁡θ(1−xR)y = x\tan \theta \left( {1 - {x \over R}} \right)

⇒10=20(1−20R)\Rightarrow 10 = 20\left( {1 - {{20} \over R}} \right)

⇒R=40\Rightarrow R = 40

40=u210⇒u=2040 = {{{u^2}} \over {10}} \Rightarrow u = 20

⇒T=20×20×1210=42s⇒t=2 s\Rightarrow T = {{20 \times 20 \times {1 \over {\sqrt 2 }}} \over {10}} = {4 \over {\sqrt 2 }}s \Rightarrow t = 2\,s

at t=2, v→=(102i^)+(102−2×10)j^t = 2,\,\overrightarrow v = \left( {10\sqrt 2 \widehat i} \right) + \left( {10\sqrt 2 - 2 \times 10} \right)\widehat j

⇒p→=10[102i^+(102−20)j^]\Rightarrow \overrightarrow p = 10\left[ {10\sqrt 2 \widehat i + \left( {10\sqrt 2 - 20} \right)\widehat j} \right]

=1002i^+(1002−200)j^= 100\sqrt 2 \widehat i + \left( {100\sqrt 2 - 200} \right)\widehat j

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