A force on an object of mass 100 g is $(10\hat{\mathrm{i}}+5\hat{\mathrm{j}})$ N. The position of that object at $\mathrm{t}=2$ s is $(\mathrm{a}\hat{\mathrm{i}}+\mathrm{b}\hat{\mathrm{j}})$ m after starting from rest. The value of $\frac{a}{b}$ will be

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Kinematics

A force on an object of mass 100 g is

(10 \hat{i} + 5 \hat{j})

N. The position of that object at

t = 2

s is

(a \hat{i} + b \hat{j})

m after starting from rest. The value of

\frac{a}{b}

will be

Answer: 2

Solution

Given:

F = 10 \hat{i} + 5 \hat{j}

N and

m = 100

= 0.1

kg.

Acceleration:

a = \frac{F}{m} = \frac{10 i ^ + 5 j ^}{0.1} = 100 \hat{i} + 50 \hat{j}

m/s²

Since the object starts from rest (

u = 0

), using the kinematic equation:

S = u t + \frac{1}{2} a t^{2} = \frac{1}{2} (100 \hat{i} + 50 \hat{j}) (2)^{2}

S = \frac{1}{2} (100 \hat{i} + 50 \hat{j}) (4) = 200 \hat{i} + 100 \hat{j}

Therefore,

a = 200

m and

b = 100

\frac{a}{b} = \frac{200}{100} = 2

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Integer answer

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