MCQEasyJEE 2023Velocity & Acceleration

JEE Physics 2023 Question with Solution

At any instant the velocity of a particle of mass 500g500 \, \text{g} is (2ti^+3t2j^)m s1\left(2t\hat{i} + 3t^2\hat{j}\right) \, \text{m s}^{-1}. If the force acting on the particle at t=1st = 1 \, \text{s} is (i^+xj^)N\left(\hat{i} + x\hat{j}\right) \, \text{N}, then the value of xx will be:

  • A

    22

  • B

    66

  • C

    33

  • D

    44

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Mass of the particle is 500g=0.5kg500 \, \text{g} = 0.5 \, \text{kg} and velocity is v=(2ti^+3t2j^)\vec{v} = \left(2t\hat{i} + 3t^2\hat{j}\right).

Find: The value of xx in the force F=(i^+xj^)N\vec{F} = \left(\hat{i} + x\hat{j}\right) \, \text{N} at t=1st = 1 \, \text{s}.

From the given velocity, acceleration is obtained by differentiating with respect to time:

a=dvdt=(2i^+6tj^)\vec{a} = \frac{d\vec{v}}{dt} = \left(2\hat{i} + 6t\hat{j}\right)

At t=1st = 1 \, \text{s},

a=(2i^+6j^)m s2\vec{a} = \left(2\hat{i} + 6\hat{j}\right) \, \text{m s}^{-2}

Now use F=ma\vec{F} = m\vec{a} with m=0.5kgm = 0.5 \, \text{kg}:

F=0.5(2i^+6j^)=(i^+3j^)N\vec{F} = 0.5\left(2\hat{i} + 6\hat{j}\right) = \left(\hat{i} + 3\hat{j}\right) \, \text{N}

Comparing with (i^+xj^)N\left(\hat{i} + x\hat{j}\right) \, \text{N}, we get x=3x = 3.

Therefore, the correct option is C.

Using Components of Force

Given: v=(2ti^+3t2j^)\vec{v} = \left(2t\hat{i} + 3t^2\hat{j}\right) and m=0.5kgm = 0.5 \, \text{kg}.

Find: The coefficient xx of the j^\hat{j}-component of force at t=1st = 1 \, \text{s}.

Differentiate each component separately:

ax=d(2t)dt=2,ay=d(3t2)dt=6ta_x = \frac{d(2t)}{dt} = 2, \qquad a_y = \frac{d(3t^2)}{dt} = 6t

So at t=1st = 1 \, \text{s},

ax=2,ay=6a_x = 2, \qquad a_y = 6

Multiply by mass to get force components:

Fx=max=0.5×2=1,Fy=may=0.5×6=3F_x = ma_x = 0.5 \times 2 = 1, \qquad F_y = ma_y = 0.5 \times 6 = 3

Hence,

F=i^+3j^\vec{F} = \hat{i} + 3\hat{j}

So the value of xx is 33.

Common mistakes

  • Differentiating the velocity incorrectly. The j^\hat{j}-component is 3t23t^2, so its derivative is 6t6t, not 3t3t or 6t26t^2. Differentiate each component with respect to tt carefully.

  • Using mass as 500500 instead of converting 500g500 \, \text{g} to 0.5kg0.5 \, \text{kg}. Force in SI units requires mass in kilograms, so always convert before applying F=ma\vec{F} = m\vec{a}.

  • Comparing the whole vector without matching components. Since F=(i^+xj^)\vec{F} = \left(\hat{i} + x\hat{j}\right), the value of xx comes from the coefficient of j^\hat{j} only. Match the i^\hat{i} and j^\hat{j} components separately.

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