Laws of Motion

JEE Main level, single correct, medium

6 of 210

Question 6 of 210

At t=0t=0, a body of mass 100100 g starts moving under the influence of a force (5i^+10j^)(5\hat{i}+10\hat{j}) N. After 22 s its position is (2xi^+5yj^)(2x\hat{i}+5y\hat{j}) m. The ratio x:yx:y is ______.

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Answer: (D) 5:45:4

Mass m=100 g=0.1 kgm = 100 \text{ g} = 0.1 \text{ kg}

Force F⃗=5i^+10j^ N\vec{F} = 5\hat{i} + 10\hat{j} \text{ N}

Acceleration a⃗=F⃗m=5i^+10j^0.1=50i^+100j^ m/s2\vec{a} = \dfrac{\vec{F}}{m} = \dfrac{5\hat{i} + 10\hat{j}}{0.1} = 50\hat{i} + 100\hat{j} \text{ m/s}^2

Since the body starts from rest, initial velocity u⃗=0\vec{u} = 0.

Position after t=2 st = 2 \text{ s} is given by r⃗=u⃗t+12a⃗t2\vec{r} = \vec{u}t + \dfrac{1}{2}\vec{a}t^2

r⃗=12(50i^+100j^)(2)2=100i^+200j^ m\vec{r} = \dfrac{1}{2}(50\hat{i} + 100\hat{j})(2)^2 = 100\hat{i} + 200\hat{j} \text{ m}

Comparing with the given position r⃗=2xi^+5yj^\vec{r} = 2x\hat{i} + 5y\hat{j}:

2x=100⇒x=502x = 100 \Rightarrow x = 50

5y=200⇒y=405y = 200 \Rightarrow y = 40

The ratio x:y=50:40=5:4x:y = 50:40 = 5:4

Answer: 5:45:4

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