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Asked in NEET 2023 · Stopping and retardation
Given: entry speed u, speed u/3 after 24 cm, uniform retardation a, final speed zero.
Idea: apply v²=u²-2as over the two stretches and divide, so the unknown retardation cancels.
First stretch: (u/3)²=u²-2a(24), so 2a(24)=u²-(u²)/9=(8u²)/9.
Remaining stretch s: 0=(u/3)²-2as, so 2as=(u²)/9.
Dividing: s/(24)=(u²)/9×9/(8u²)=1/8, so s=3 cm.
Total penetration =24+3=27 cm.
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