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Analysis of Kinematic and Dynamic Vectors of a Particle

The position of an object having mass 0.1 kg0.1 \text{ kg} as a function of time tt is given as r=(10t2i^+5t3j^) m\vec{r} = \left(10t^2 \hat{i} + 5t^3 \hat{j}\right) \text{ m}. At t=1 st = 1 \text{ s}, which of the following statements are correct?

A. The linear momentum p=(2i^+1.5j^) kgm/s\vec{p} = \left(2 \hat{i} + 1.5 \hat{j}\right) \text{ kg}\cdot\text{m/s}.

B. The force acting on the object F=(2i^+3j^) N\vec{F} = \left(2 \hat{i} + 3 \hat{j}\right) \text{ N}.

C. The angular momentum of the object about its origin L=15k^ J s\vec{L} = 15 \hat{k} \text{ J s}.

D. The torque acting on the object about its origin τ=20k^ N m\vec{\tau} = 20 \hat{k} \text{ N m}.

Choose the correct answer from the options given below :

Options

A

A, B and C only

B

B, C and D only

C

A, C and D only

D

A, B and D only

Correct

Step-by-Step Solution

Given mass m=0.1 kgm = 0.1\text{ kg} and position vector r(t)=10t2i^+5t3j^ m\vec{r}(t) = 10t^2\hat{i} + 5t^3\hat{j}\text{ m}.

Differentiating r(t)\vec{r}(t) with respect to time yields velocity v(t)=20ti^+15t2j^ m/s\vec{v}(t) = 20t\hat{i} + 15t^2\hat{j}\text{ m/s} and acceleration a(t)=20i^+30tj^ m/s2\vec{a}(t) = 20\hat{i} + 30t\hat{j}\text{ m/s}^2. At t=1 st = 1\text{ s}, the values are r=10i^+5j^\vec{r} = 10\hat{i} + 5\hat{j}, v=20i^+15j^\vec{v} = 20\hat{i} + 15\hat{j}, and a=20i^+30j^\vec{a} = 20\hat{i} + 30\hat{j}.

  1. Linear momentum: p=mv=0.1(20i^+15j^)=2i^+1.5j^ kgm/s\vec{p} = m\vec{v} = 0.1\left(20\hat{i} + 15\hat{j}\right) = 2\hat{i} + 1.5\hat{j}\text{ kg}\cdot\text{m/s} (Statement A is correct).
  2. Force: F=ma=0.1(20i^+30j^)=2i^+3j^ N\vec{F} = m\vec{a} = 0.1\left(20\hat{i} + 30\hat{j}\right) = 2\hat{i} + 3\hat{j}\text{ N} (Statement B is correct).
  3. Angular momentum: L=r×p=(10i^+5j^)×(2i^+1.5j^)=(1510)k^=5k^ J s\vec{L} = \vec{r} \times \vec{p} = (10\hat{i} + 5\hat{j}) \times (2\hat{i} + 1.5\hat{j}) = (15 - 10)\hat{k} = 5\hat{k}\text{ J s} (Statement C is incorrect).
  4. Torque: τ=r×F=(10i^+5j^)×(2i^+3j^)=(3010)k^=20k^ N m\vec{\tau} = \vec{r} \times \vec{F} = (10\hat{i} + 5\hat{j}) \times (2\hat{i} + 3\hat{j}) = (30 - 10)\hat{k} = 20\hat{k}\text{ N m} (Statement D is correct).

Therefore, statements A, B, and D only are correct.

Correct Option: D

Analysis of Kinematic and Dynamic Vectors of a Particle | Physics PYQ Solution - JEE Challenger