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Question

A bullet fired into a fixed wooden block loses half of its velocity after penetrating 40cm. It comes to rest after penetrating a further distance of


A

403cm

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B

203cm

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C

225cm

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D

265cm

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Solution

The correct option is A

403cm


Step 1: Given data

Distance, s=40cm=0.4m

Final velocity, v1 at distance, 0.4mis v1=u2, where, uis the initial velocity.

Step 2: Formula used

Equation of motion,

v2=u2+2as

Where, vis the final velocity, uis the initial velocity, a is the deceleration, s is the distance travelled.

Step 3: Find the deceleration a in terms of the final velocity v

Substituting, s=0.4m

v12=u2-2asa=u2-u222sa=4u2-u24×2sa=3u24×2×0.4a=3u23.2a=15u216........(i)

Step 4: Find the distance travelled by the bullet after it reaches half the initial speed

Using (i), putting the value of a

Let distance after further penetrating 40cm is x till it come to rest such that the final velocity, v2=0.

The initial velocity will be, u'=u2

Again using the equation of motion,

v2=u2-2as

v22=u'2-2ax

x=(u'2-v22)2a

x=u22-022×15u216x=403cm

Thus, the distance travelled by the bullet in the wooden block is 403cm.

Hence, option A is the correct option.


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