Question

If a body losses half of its velocity on penetrating 3 cm in wooden block , then how much will it penetrate more before coming to rest ? [AIEEE 2002]​

Answer 1

• s= 3cm
• v= u/2 cm/s

Applying third law of motion,

${v}^{2} - {u}^{2} = 2as \\ \implies \: \dfrac{ {u}^{2} }{4} - {u}^{2} = 2a \times 3 \\ \implies \: - 3{u}^{2} = 6a \\ \implies \: a = {u}^{2} \div (- 2)$

When

• v= 0m/s

$a = {u}^{2} \div (- 2)$

${v}^{2} - {u}^{2} = 2as \\ \implies \: - {u}^{2} = 2 \times ( \dfrac{1}{ - 2 {u}^{2}}) \times s \\ \implies \: s = - {u}^{2} \div ( \dfrac{1}{ - 2 {u}^{2} } \times{1/2} ) \\ \implies \: s = {4 cm$

Answer 2

Explanation :- Let initial velocity of body at point A be V , AB is 3cm

• From v² = u²-2as
• (v/2)² = v² -2as
• a = v²/8

let on penetrating 3cm in a wooden blocks , the body moves X distance from B To C

• so, for B to C , u = v/2 ,v = 0
• s = x , a = v²/8 [Deceleration]
• (0)² = (v/2)² -2•v²/8•x
• x = 1cm

Note :- Here, it is assumed that retardation is uniform

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