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Laws of Motion question

2022 · 25 Jun · Shift 1 · Q70
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  5. /2022 · 25 Jun · Shift 1 · Q70

Laws of Motion question

2022 · 25 Jun · Shift 1 · Q70

JEE MainPhysicsLaws of MotionNumerical+4 / −1
A force on an object of mass 100 g is (10i^+5j^)\left( {10\widehat i + 5\widehat j} \right)(10i+5j​) N. The position of that object at t = 2 s is (ai^+bj^)\left( {a\widehat i + b\widehat j} \right)(ai+bj​) m after starting from rest. The value of ab{a \over b}ba​ will be ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 2

  1. Given data
  • Mass of the object: 100 g=0.1 kg100\text{ g} = 0.1\text{ kg}100 g=0.1 kg
  • Force on the object: F⃗=10i^+5j^  N\vec F = 10\hat i + 5\hat j \;\text{N}F=10i^+5j^​N
  • Initial velocity: object starts from rest, so u⃗=0\vec u = 0u=0
  • Time: t=2 st=2\text{ s}t=2 s
  1. Find acceleration using Newton's second law

Using F⃗=ma⃗\vec F = m\vec aF=ma we get a⃗=F⃗m=10i^+5j^0.1\vec a = \frac{\vec F}{m} = \frac{10\hat i + 5\hat j}{0.1}a=mF​=0.110i^+5j^​​

So, a⃗=100i^+50j^  m/s2\vec a = 100\hat i + 50\hat j \;\text{m/s}^2a=100i^+50j^​m/s2

  1. Find displacement after 2 s

Since initial velocity is zero, r⃗=u⃗t+12a⃗t2=12a⃗t2\vec r = \vec u t + \frac{1}{2}\vec a t^2 = \frac{1}{2}\vec a t^2r=ut+21​at2=21​at2

At t=2t=2t=2 s, r⃗=12(100i^+50j^)(2)2\vec r = \frac{1}{2}(100\hat i + 50\hat j)(2)^2r=21​(100i^+50j^​)(2)2

r⃗=12(100i^+50j^)⋅4\vec r = \frac{1}{2}(100\hat i + 50\hat j)\cdot 4r=21​(100i^+50j^​)⋅4

r⃗=2(100i^+50j^)\vec r = 2(100\hat i + 50\hat j)r=2(100i^+50j^​)

r⃗=200i^+100j^\vec r = 200\hat i + 100\hat jr=200i^+100j^​

Hence, a=200,b=100a=200,\qquad b=100a=200,b=100

  1. Calculate ab\dfrac{a}{b}ba​

ab=200100=2\frac{a}{b} = \frac{200}{100} = 2ba​=100200​=2

  1. Comparison with stored answer

Derived answer is 222, which matches the stored correct answer.

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