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Physics Question 28 – JEE-MAIN 2026

At t=0, a body of mass 100 g starts moving under the influence of a force (5i^+10j^) N. After 2 s its position is (2xi^+5yj^) m. The ratio x:y is _______.

Recall Newton's second law to find the acceleration of the body and the kinematic equation for position.

Step 1: Calculate the acceleration of the body✦ Active

First, convert the mass to kilograms and then use Newton's second law, F=ma, to find the acceleration vector.

m=100 g=0.1 kg F=(5i^+10j^) N a=Fm=(5i^+10j^) N0.1 kg=(50i^+100j^) m/s2
Step 2: Determine the final position of the body○ Expand

Assuming the body starts from rest at the origin (initial position r0=0 and initial velocity v0=0), use the kinematic equation for position at time t=2 s.

r=r0+v0t+12at2 r=0+0(2)+12(50i^+100j^)(2)2 r=12(50i^+100j^)(4)=(100i^+200j^) m
Step 3: Equate components and find the ratio x:y○ Expand

The calculated final position is (100i^+200j^) m. This is given as (2xi^+5yj^) m. Equate the corresponding components to solve for x and y, then find their ratio.

(2xi^+5yj^)=(100i^+200j^) 2x=100x=50 5y=200y=40 x:y=50:40=5:4
💡 Teacher's Secret Hint

Ensure units are consistent (SI units) before performing calculations.

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