The projectile motion of a particle of mass 5 g is shown in the figure.

The initial velocity of the particle is $$5\sqrt{2}$$ ms$$^{-1}$$ and the air resistance is assumed to be negligible. The magnitude of the change in momentum between the points A and B is $$x \times 10^{-2}$$ kgms$$^{-1}$$. The value of $$x$$, to the nearest integer, is ___.

Upon reviewing the previous derivation, a typo was noticed in the mathematical formatting line of Section 3. Here is the corrected, complete solution:

1. Identify the Projectile Parameters

From the problem statement , we are given:

• Mass of the particle ($$m$$) = $$5\text{ g} = 5 \times 10^{-3}\text{ kg}$$
• Initial velocity magnitude ($$u$$) = $$5\sqrt{2}\text{ m s}^{-1}$$
• Angle of projection ($$\theta$$) = $$45^\circ$$

2. Analyze Velocity Components at Points A and B

Let's break down the velocity of the particle into horizontal ($$x$$) and vertical ($$y$$) components at both positions:

• At Launch Point A:
  The particle is moving upward and to the right:

  $$\vec{v}_A = (u \cos\theta)\hat{i} + (u \sin\theta)\hat{j}$$

• At Landing Point B:
  Since air resistance is negligible, the horizontal velocity remains unchanged. Due to symmetry, the vertical component retains its magnitude but points downward:

  $$\vec{v}_B = (u \cos\theta)\hat{i} - (u \sin\theta)\hat{j}$$

3. Calculate the Change in Momentum ($$\Delta \vec{p}$$)

The change in velocity ($$\Delta \vec{v}$$) between point $$A$$ and point $$B$$ is:

$$\Delta \vec{v} = \vec{v}_B - \vec{v}_A$$

$$\Delta \vec{v} = \left[(u \cos\theta)\hat{i} - (u \sin\theta)\hat{j}\right] - \left[(u \cos\theta)\hat{i} + (u \sin\theta)\hat{j}\right]$$

$$\Delta \vec{v} = -2u \sin\theta \hat{j}$$

The magnitude of this change in velocity is simply $$2u \sin\theta$$. Therefore, the magnitude of the change in linear momentum is:

$$|\Delta \vec{p}| = m \cdot |\Delta \vec{v}| = 2mu \sin\theta$$

4. Solve for $$x$$

Substitute the given values into the momentum equation:

$$|\Delta \vec{p}| = 2 \times (5 \times 10^{-3}\text{ kg}) \times (5\sqrt{2}\text{ m s}^{-1}) \times \sin 45^\circ$$

Since $$\sin 45^\circ = \frac{1}{\sqrt{2}}$$, the radical terms cancel out perfectly:

$$|\Delta \vec{p}| = 2 \times 5 \times 10^{-3} \times 5\sqrt{2} \times \frac{1}{\sqrt{2}}$$

$$|\Delta \vec{p}| = 2 \times 5 \times 5 \times 10^{-3}$$

$$|\Delta \vec{p}| = 50 \times 10^{-3} = 5 \times 10^{-2}\text{ kg m s}^{-1}$$

The problem states that the magnitude of the change in momentum is written in the form $$x \times 10^{-2}\text{ kg m s}^{-1}$$. Comparing both expressions:

$$x = 5$$

Conclusion

The corrected value of $$x$$, to the nearest integer, is 5.