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 [JEE Adv 2024 P1]

Correct Answer is : 0.5 m,−5 kg m/s0.5\,{m}, -5\,{kg}\,{m}/{s}0.5m,−5kgm/s

Correct Answer is : 0.5 m,−5 kg m/s0.5\,{m}, -5\,{kg}\,{m}/{s}0.5m,−5kgm/s

Test Index

Impulse & Momentum

Section: Physics

Question : 2 of 16

Marks: +1, -0

x

F=(-20 x+10)\,\text{N}

x

t=0\,\text{s}

x=1\,\text{m}

t=\frac{\pi}{4}\,\text{s}

Solution:

Equation of motion:

m\ddot{x}=F=(-20x+10)

Put

m=5\,\text{kg}

5\ddot{x}=-20x+10\Rightarrow \ddot{x}=-4x+2=-4\left(x-\frac{1}{2}\right)

Let

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

Then

\ddot{X}=-4X

, so the angular frequency is

\omega=2\,\text{rad/s}

Initial conditions: at

t=0

x=1\,\text{m}

and velocity

v=0

Thus

X(0)=\frac{1}{2}

and

\dot X(0)=0

The solution is

X(t)=\frac{1}{2}\cos(2t)

Hence

x(t)=\frac{1}{2}+\frac{1}{2}\cos(2t)=\cos^2 t

t=\frac{\pi}{4}\,\text{s}

x=\cos^2\left(\frac{\pi}{4}\right)=\frac{1}{2}=0.5\,\text{m}

Velocity is

v=\dot{x}=-\sin(2t)

t=\frac{\pi}{4}\,\text{s}

v=-\sin\left(\frac{\pi}{2}\right)=-1\,\text{m/s}

Momentum is

p=mv=5(-1)=-5\,\text{kg}\,\text{m}/\text{s}

Therefore position and momentum are

0.5\,\text{m}

and

-5\,\text{kg}\,\text{m}/\text{s}

This matches option (2).

Go to Question:

Prev Question

Next Question