Question

A bullet fired into a fixed target loses half of its velocity after penetrating 3cm . How much further it will penetrate before coming to rest , assuming that it faces constant Resistance to motion ? ​

Answer 1

Given:-

Case :- 1

• Distance travelled = 3cm.

• Initial Velocity = u

• Final Velocity = $\sf{\dfrac{u}{2}}$.

Case :- 2

• Distance travelled = x

• Initial Velocity = $\sf{\dfrac{u}{2}}$.
• Final Velocity = 0m/s

To Find:-

• The Distance bullet will penetrate before coming to rest.

Formulae used:-

• v² - u² = 2as

Where,

• v = Final Velocity
• u = Initial Velocity
• a = Accceleration
• s = Distance

Now, Case 1

→ v² - u² = 2as

→ $\sf{ (\dfrac{u}{2})^2 - u^2 = 2 * a * 3}$

→ $\sf{ \dfrac{u^2}{4} - u^2 = 6a}$

→ $\sf{ \dfrac{-3u^2}{4} = 6a}$

→ $\sf{ a = \dfrac{-u^2}{8}}$......1

Now, In Case 2

→ v² - u² = 2as

→ $\sf{0 - (\dfrac{u}{2})^2 = 2 * \dfrac{-u^2}{8} * x}$......using 1

→ $\sf{ \dfrac{u^2}{4} = \dfrac{-2u^2}{8} * x}$

→ $\sf{ -8u^2 = -8u^2 * x}$

→ $\sf{ \dfrac{-8u^2}{-8u^2} = x}$

→ $\sf{ x = 1cm}$.

Hence, The bullet will penetrate 1cm more.

Answer 2

method :- Using Laws of motion

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

From v² = u² -2as

⟹ (v/2)² = v² -2a* 3

⟹ 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
• v² /4 = v²/4 × x
• x = 1 cm Answer .

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Method:-02

Explanation :- Using work- energy theorem

Total work done = change in kinetic energy W = ∆k

case I

• - F × 3 =1/2 m (v₀/2)² - 1/mv₀²
• Where , F is resistive force and V₀ is initial speed

case II

• Let the further distance travelled by the bullet before coming to rest be s .

• ∴ -1/8 mv₀² (3+s) = -1/2mv₀²
• ⟹ 1/4 (3+s) = 1
• ⟹ 3/4+s/4 = 1
• ⟹s/4 = 1-3/4
• ⟹s/4 = 1/4
• ⟹s = 1cm

∴s = 1 cm Answer