Question

a bullet loses 25% of its speed while travelling 14cm inside a wooden blocks. How much further will it travel before stopping

Answer 1

Here Let the Intial velocity of Bullet be v m/s
after traveling 14 cm the velocity becomes 75v/100 = 3v/4

Then by work-energy theorem,  W = (1/2)m{ v² - 9v²/16 }=> F×s = (1/2)m{ 7v²/16}=> F×(14/100)= 7mv²/32=> F = 25mv²/16


Distance travelled by the bullet before it stops=> F × s = mv²/2
=> 25mv²/(16) × s = mv²/2=> s = 8/25 m = 32 cms
So the distance traveled by bullet before stopping is 32 cms
So the distance bullet has to further travel after 14 cms to stop is 18 cms.

Answer 2

Let initial velocity of bullet is u ,
A/C to question,
bullet loses25% of its speed while traveling inside a wooden blocks.
means final velocity = u -25 % of u = u - u/4 = 3u/4

Now, use formula,
v² = u² + 2aS
Here, v = 3u/4, S = 14cm
∴ (3u/4)² = u² + 2a(14)
⇒9u²/16 - u² = 28a
⇒ - 7u²/16 = -28a
⇒ a = -u²/64

Let total distance covered by bullet before stopping is x = 14 + P
where P is distance covered by bullet after covering 14cm.
Now, use formula, v² = u² + 2as
here, v = 0, a = -u²/64
∴ 0 = u² - 2(u²/64)x
x = 32 cm
So, P = 32 - 14 = 18cm

Hence, bullet further travel 18cm before stopping.