Laws of Motion

JEE Main level, numerical answer, medium

32 of 210

Question 32 of 210

A force on an object of mass 100g100g is (10i^+5j^)N\left(10\hat{i}+5\hat{j}\right)N. The position of that object at t=2st=2s is (ai^+bj^)m\left(a\hat{i}+b\hat{j}\right)m after starting from rest. The value of ab\dfrac{a}{b} will be _____ .

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Answer: 2

a⃗=F⃗m=10i^+5j^0.1=100i^+50j^\vec{a}=\dfrac{\vec{F}}{m}=\dfrac{10\hat{i}+5\hat{j}}{0.1}=100\hat{i}+50\hat{j}
Displacement covered in vector form can be written as, s⃗=u⃗t+12a⃗t2\vec{s}=\vec{u}t+\dfrac{1}{2}\vec{a}{t}^{2}
⇒s⃗=12(100i^+50j^)(2)2\Rightarrow \vec{s}=\dfrac{1}{2}\left(100\hat{i}+50\hat{j}\right){\left(2\right)}^{2}
, ⇒s⃗=200i^+100j^\Rightarrow \vec{s}=200\hat{i}+100\hat{j}
Comparing it with, s⃗=ai^+bj^\vec{s}=a\hat{i}+b\hat{j}, we get a=200\text{&}b=100
Therefore, ab=2\dfrac{a}{b}=2

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