Question

A body of mass 2 kg starts moving from the origin (0, 0) with an initial velocity of \( (4\hat{i} + 4\hat{j})\ \text{ms}^{-1} \). A constant force of \( -20\hat{j} \ \text{N} \) is applied on the body. When the Y-coordinate of the position of the body becomes zero again, then its X-coordinate is

• A. 3.2 m
• B. 4.2 m
• C. 2.4 m
• D. 2.8 m

Hint:

Treat the x and y directions independently; vertical motion here behaves like a projectile under constant acceleration.

Answer 1

The Correct Option is A

Step 1: Force \( \vec{F} = -20\hat{j} \), mass \( m = 2\ \text{kg} \) \[ a_y = \frac{F_y}{m} = \frac{-20}{2} = -10\ \text{ms}^{-2} \] Step 2: Time for vertical motion to return to y = 0: \[ s_y = u_yt + \frac{1}{2}a_yt^2 \Rightarrow 0 = 4t - 5t^2 \Rightarrow t = 0, t = 0.8\ \text{s} \] Step 3: x-displacement during this time: \[ x = u_xt = 4 \times 0.8 = 3.2\ \text{m} \]