JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Conservation of Momentum

A projectile is thrown from a point O on the ground at an angle 45° from the vertical and with a speed

52

m/s. The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, 0.5 s after the splitting. The other part, t seconds after the splitting, falls to the ground at a distance x meters from the point O. The acceleration due to gravity

g = 10

m/s

^{2}

.

The value of t is ____.

Answer: 0.5

Solution

Initial velocity components:

u_{x} = 5

m/s,

u_{y} = 5

m/s (45° from vertical means 45° from horizontal).

Time of flight:

T = \frac{2 u _{y}}{g} = \frac{2 \times 5}{10} = 1

s. Time to reach highest point = T/2 = 0.5 s.

At the highest point, velocity is purely horizontal:

v_{x} = 5

m/s.

When the projectile splits, one part falls vertically (taking 0.5 s), so it was at height

h = \frac{1}{2} g t^{2} = \frac{1}{2} \times 10 \times (0.5)^{2} = 1.25

m.

By momentum conservation (horizontal):

2 m \times 5 = m \times 0 + m \times v

, so

v = 10

m/s for the other part.

The other part also takes the same time to fall from the same height: t = 0.5 s.

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Numerical answer

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JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Conservation of Momentum

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