JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Conservation of Momentum

The projectile motion of a particle of mass

5 g

is shown in the figure. The initial velocity of the projectile is

52 ms^{- 1}

and air resistance is assumed to be negligible. The magnitude of change in momentum between the points A to B is

x \times 1 0^{- 2} kg m s^{- 1}

.

The value of x, to the nearest integer, is ____

Answer: 5

Solution

The projectile leaves A at

4 5^{\circ}

with speed

52 m s^{- 1}

, so its components there are

v_{x} = 52 cos 4 5^{\circ} = 5 m s^{- 1}

and

v_{y} = 5 m s^{- 1}

. B lies at the same level, so at B the horizontal component is unchanged and the vertical one is reversed. Only the vertical momentum changes, and with

m = 5 g = 5 \times 1 0^{- 3} kg

∣ Δ p ∣ = m ∣ v_{y, B} - v_{y, A} ∣ = 5 \times 1 0^{- 3} \times 10 = 5 \times 1 0^{- 2} kg m s^{- 1}

x = 5

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Numerical answer

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JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Conservation of Momentum

Solution

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