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JEE Main 2022 · 27 Jul · Shift 2, Physics question 32

A body of mass 10 kg is projected at an angle of 45° with the horizontal. The trajectory of the body is observed to pass through a point (20, 10). If T is the time of flight, then its momentum vector, at time t=T2t=\dfrac T{\sqrt2}, is ______.

[Take g = 10 m/s2^2]

Choose one option
Answer and worked solution

Official answer: D

Speed. At 45°, y=x−gx2u2y=x-\frac{gx^2}{u^2}; through (20,10)(20,10) this gives u=20u=20 m/s, so each component is 10210\sqrt2.

Time. T=22T=2\sqrt2 s and t=2t=2 s, when vy=102−20v_y=10\sqrt2-20.

Momentum. p⃗=1002 i^+(1002−200)j^\vec p=100\sqrt2\,\hat i+\left(100\sqrt2-200\right)\hat j.

(D)

Topics:Projectile motionImpulse and momentum
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