JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Kinematics

A body of mass 10 kg is projected at an angle of

4 5^{\circ}

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 = \frac{T}{2}

, is ____
[Take

g = 10 m / s^{2}

]

Answer: (D)

Solution

Using the trajectory equation

y = x tan θ - \frac{g x ^{2}}{2 u ^{2} c o s ^{2} θ}

with

θ = 4 5^{\circ}

, substituting point (20,10):

10 = 20 - \frac{10 \times 400}{2 u ^{2} \times 0.5}

, solving gives

u = 20

m/s. Time of flight

T = \frac{2 u s i n θ}{g} = \frac{2 \times 20}{2 \times 10} = 22

s. At

t = \frac{T}{2} = 2

s, velocity components are

v_{x} = u cos 4 5^{\circ} = 102

m/s (constant) and

v_{y} = u sin 4 5^{\circ} - g t = 102 - 20

m/s. Momentum

p = m v = 10 [102 \hat{i} + (102 - 20) \hat{j}] = 1002 \hat{i} + (1002 - 200) \hat{j}

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Single correct option