Question

A bullet loses one fourth of its velocity after penetrating through a sand bag. How
many identical sand bags are required to stop the bullet?
(A) 9
(B) 7
(C) 5
(D) 3

Answer 1

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Given, Initial velocity, U=v

Final velocity, V=2v

Distance, s=20cm

Let, the further distance of penetration before it comes to rest be x.

V2=U2−2as

(2V)2=V2−2a×20

40a=V2−4V2

40a=43V2  ...(i)

and V2=u2+2as

v2=0+2a×(20+x)

v2=2×1603v2×(20+x)   [From Eq. (i)]

1=803(20+x)

380=20+x

x=380−20

x=380−60

x=320=6.66cm

Answer 2

Given : A bullet loses one fourth of its velocity after penetrating through a sand bag.

To Find : How many identical sand bags are required to stop the bullet?

(A) 9

(B) 7

(C) 5

(D) 3

Solution:

Assume that initial Velocity = 4v m/s

bullet loses one fourth of its velocity after penetrating through a sand bag.

Hence  velocity after penetrating through a sand bag. = 4v - (1/4)4v  = 3v m/s

V² - U² = 2aS

S = distance covered in one sand bag

=> (3v)² - (4v)² = 2aS

=> -7v² = 2aS

let say n sand bags used then S = nS

final velocity = 0

=> 0 - (4v)² = 2a(nS)

=>  -16v² = 2a(nS)

on dividing

=> 7/16   = 1/n

=> n = 16/7

=> n = 2.29

Hence 3  Sand bags required

Correct option is D)  3

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