Question

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 ________.

Hint:

In symmetric projectile motion, the change in momentum is always vertical and equals \(2mv\sin\theta\).

Answer 1

Correct Answer: 5

Step 1: Analyze the velocity at launch (A) and landing (B). In a standard symmetric projectile motion, the horizontal velocity remains constant (\(v_x = v\cos\theta\)), while the vertical velocity reverses direction (\(v_y\) at A is \(v\sin\theta\); at B it is \(-v\sin\theta\)).
Step 2: Calculate the change in velocity (\(\Delta \vec{v}\)). \[ \Delta \vec{v} = \vec{v}_B - \vec{v}_A = (v_x \hat{i} - v_y \hat{j}) - (v_x \hat{i} + v_y \hat{j}) = -2v_y \hat{j} \] Assuming a standard \(45^\circ\) launch angle for the given \(5\sqrt{2}\) magnitude: \[ v_y = 5\sqrt{2} \times \sin(45^\circ) = 5\sqrt{2} \times \frac{1}{\sqrt{2}} = 5 \text{ ms}^{-1} \] \[ |\Delta \vec{v}| = 2 \times 5 = 10 \text{ ms}^{-1} \]
Step 3: Calculate change in momentum (\(\Delta P\)). \[ \Delta P = m \times |\Delta v| = (5 \times 10^{-3} \text{ kg}) \times 10 \text{ ms}^{-1} = 5 \times 10^{-2} \text{ kgms}^{-1} \] Comparing with \(x \times 10^{-2}\), we get \(x = 5\).