Question

5|M
121. A bullet fired into a target, lose half of its velocity afte
penetrating 25 cm. How much further it will penetrate
before coming to rest
(a) V25cm (b) 25cm (c) 75cm (d) 8.3cm
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Answer 1

Answer:

Answer:

Let the velocity of the bullet when it strikes the target be v.

Velocity after the bullet penetrated 25 cm =v/2.

Using v²=u²+2aS

(v/2)²=v²+2a × 25

v²/4=v²+ 50a

-3v²/4= 50a

⟹ a = -3v²/200

Since the resistance offered by the material to the bullet is same throughout, a remains constant.

Using v²=u²+2aS for motion after 25 cm

Finally the bullet is at rest, hence v=0

Initial velocity for motion after 25 cm =v/2

v²=u²+2aS

0²=(v/2)²+2 × -3v²/200× S

-3v²/100× S = - v²/4

S = 1/4 × (100/3) = 25/3 = 8.3

So the required distance = 8.3 cm

So the required answer is (d) 8.3cm

Answer 2

Answer:

option D is right answer

Let the velocity of the bullet when it strikes the target be v.

Velocity after the bullet penetrated 15cm =v2.

Using v²=u²+2aS

(v2)2=v2+2a(15)

⟹v24=v2+30a

⟹a=−v240

Since the resistance offered by the material to the bullet is same throughout, a remains constant.

Using v²=u²+2aS for motion after 15cm

Finally the bullet is at rest, hence v=0

Initial velocity for motion after 15cm =v2

⟹(0)2=(v2)2+2(−v240)(S)

⟹2(v240)(S)=(v2)2

⟹v220S=v24

⟹S=5cm