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

(A) $-0.5 \mathrm{~m}, 5 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$

(B) $0.5 \mathrm{~m}, 0 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$

(D) $-1 \mathrm{~m}, 5 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$

JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Elasticity & SHM

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

Answer: (C)

The force can be written as

F = - 20 (x - \frac{1}{2}) = - 20 X

where

X = x - \frac{1}{2}

. This represents SHM about

x = 0.5

m with

ω = 20/5 = 2

rad/s, giving period

T = π

s. At

t = 0

, the particle is at rest at

x = 1

m, which is an extreme position (amplitude

A = 0.5

m). In time

t = π /4

s, the phase covered is

ω t = π /2

(quarter period), so the particle reaches the equilibrium position

x = 0.5

m with maximum speed

v = ω A = 2 \times 0.5 = 1

m/s toward negative x. Momentum

= 5 \times (- 1) = - 5

kg m/s.

JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Single correct option