Question

If a body loses half of its velocity on penetrating 3 cm in a wooden block the how much will it penetrate more before coming to rest ?

Answer 1

Answer:

hi friend

Explanation:

u be the initial velocity and ( 3+d) is the distance when it stops.

u^2 = 2a ( 3+d) ————-1

At 3 cm from start , its speed is u/2 And it moves a distance of d cm and stops.

Therefore

(u/2)^2 = 2a d ————2

(1) % (2) gives

4 = (3+d) / d

d = 1 cm.

Suppose the initial velocity is ‘v'

And so the final velocity after covering 3 cm will be v/2

So by equation of motion,

v^2= u^2 +2as

(v/2)^2= v^2 +2as

v^2/4 -v^2= 2as

2as= -3v^2/4

a= -3v^2/4×2s

a= -3v^2/24

a= -v^2/8

Now,

v^2= u^2 + 2as

And the final velocity is zero

0= v^2/4 -2×(v^2/8)×s

s= (v^2/4)/2× v^2/8

So,

s= 1cm

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Answer 2

Let the velocity = v = u/2

Using third equation of Motion.

v² = u² - 2aS

(u/2)² = u² - 2×3×a

u²/4 - u² = -6a

-3u²/4 = -6a

a = 3u²/24

Again,

When body has penetrated then its final velocity becomes zero so that It will come to rest.

v² = (u/2)² - 2aS'

u²/4 = 2×3u²/24 × S'

u²/4 = u²S'/4

u² = u²S'

S' = 1 cm.

Hence,

Before coming to rest it will penetrate = S' = 1 cm.