Let a bullet is fired into a wooden block and after penetrating 90cm into the block, the bullet loses half of its velocity. If the further distance travelled by the bullet before coming to the rest (in m ) is 0.3x then find x .

Question

Let a bullet is fired into a wooden block and after penetrating 90cm into the block, the bullet loses half of its velocity. If the further distance travelled by the bullet before coming to the rest (in m ) is 0.3x then find x . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Using v2-u2=2as1 ((u/2))2-u2=2as1 ⇒ a=(- 3 u2/8 s1) For 2 textnd part of motion, (0-((u/2))2=2 as 2. Substituting for 'a' we get, s2=(s1/3)=(90/3)cm=30cm=0.3m

Q. Let a bullet is fired into a wooden block and after penetrating $90 c m$ into the block, the bullet loses half of its velocity. If the further distance travelled by the bullet before coming to the rest (in $m$ ) is $0.3 x$ then find $x$ .

Using $v^{2} - u^{2} = 2 a s_{1}$
$(\frac{u}{2})^{2} - u^{2} = 2 a s_{1}$
$\Rightarrow a = \frac{- 3 u ^{2}}{8 s _{1}}$
For $2^{nd}$ part of motion,
$(0 - (\frac{u}{2})^{2} = 2 a s_{2}$
Substituting for $^{'} a^{'}$ we get,
$s_{2} = \frac{s _{1}}{3} = \frac{90}{3} c m = 30 c m = 0.3 m$

Q. Let a bullet is fired into a wooden block and after penetrating 90cm into the block, the bullet loses half of its velocity. If the further distance travelled by the bullet before coming to the rest (in m ) is 0.3x then find x .

Using v2−u2=2as1​ (2u​)2−u2=2as1​ ⇒a=8s1​−3u2​ For 2nd part of motion, (0−(2u​)2=2as2​ Substituting for ′a′ we get, s2​=3s1​​=390​cm=30cm=0.3m

Q. Let a bullet is fired into a wooden block and after penetrating $90 c m$ into the block, the bullet loses half of its velocity. If the further distance travelled by the bullet before coming to the rest (in $m$ ) is $0.3 x$ then find $x$ .

Using $v^{2} - u^{2} = 2 a s_{1}$
$(\frac{u}{2})^{2} - u^{2} = 2 a s_{1}$
$\Rightarrow a = \frac{- 3 u ^{2}}{8 s _{1}}$
For $2^{nd}$ part of motion,
$(0 - (\frac{u}{2})^{2} = 2 a s_{2}$
Substituting for $^{'} a^{'}$ we get,
$s_{2} = \frac{s _{1}}{3} = \frac{90}{3} c m = 30 c m = 0.3 m$