JEE Advanced 2024 — Paper 1 — PHY-I solution

JEE Advanced2024Paper 1PHY-I

A block of mass 5 kg moves along the $x$ -direction subject to the force $F = (- 20 x + 10)$ N, with the value of $x$ in metre. At time $t = 0$ s, it is at rest at position $x = 1$ m. The position and momentum of the block at $t = (π /4)$ s are

A
$- 0.5$ m, $5$ kg m/s
B
$0.5$ m, $0$ kg m/s
C
$0.5$ m, $- 5$ kg m/s
D
$- 1$ m, $5$ kg m/s

Solution

The force can be rewritten as $F = - 20 (x - 0.5)$ , which is a linear restoring force about $x_{0} = 0.5$ m. This is simple harmonic motion with

$ω^{2} = \frac{20}{5} = 4 \Rightarrow ω = 2$ rad/s.

Let $X = x - 0.5$ . At $t = 0$ : $X_{0} = 0.5$ m and $\dot{X}_{0} = 0$ . So $X (t) = 0.5 cos (2 t)$ and $\dot{X} (t) = - sin (2 t)$ .

At $t = π /4$ s: $2 t = π /2$ , hence $X = 0$ and $\dot{X} = - 1$ m/s.

Position: $x = 0.5 + 0 = 0.5$ m.

Momentum: $p = m \overset{x}{˙} = 5 \times (- 1) = - 5$ kg m/s.

$∴$ Option (C) is correct.

← Back to full paper

Practice more PHY-I

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →