Question

A bullet enters a wooden block with speed Vo. its speed becomes half after travelling a distance of 6cm, how much distance will it move before it stops?

Answer 1

Answer:

The bullet will cover a distance of 8 cm inside the wooden block before coming to rest.

Therefore it will travel 2 cm further

Explanation:

The bullet enters with velocity $v_0$ m/s after travelling a distance of 0.06 m (6 cm) its velocity becomes $\frac{v_0}{2}$ m/s.

now we can clearly see that

initial velocity  $(u) = v_0$ m/s

final velocity $(v) = \frac{v_0}{2}$ m/s

distance traveled $(s) = 0.06$ m

Now using 3rd equation of motion

$v^2 = u^2 + 2as\\(\frac{v_0}{2})^2 = (v_0)^2 + 2a*0.06\\\frac{(v_0)^2}{4} = (v_0)^2 + a*0.12\\(v_0)^2 = 4(v_0)^2 + a*0.48\\-3(v_0)^2 = + a*0.48\\a = \frac{-3(v_0)^2}{0.48} = \frac{-(v_0)^2}{0.16}$

clearly it is seen that the velocity is decreasing

now let us calculate distance traveled by the bullet before it comes to rest

$Using \ 1st \ equation \ of \ motion\\v = u+at\\0 = v_0 - \frac{(v_0)^2}{0.16}t\\-v_0=- \frac{(v_0)^2}{0.16}t\\1=\frac{v_0}{0.16}t\\\\t = \frac{0.16}{v_0}$

$Using \ 2nd \ equation \ of \ motion\\s = ut + 0.5at^2\\s = v_0*\frac{0.16}{v_0} - \frac{1}{2}*\frac{(v_0)^2}{0.16}*(\frac{0.16}{v_0})^2\\s = 0.16-\frac{1}{2}*0.16\\s = 0.16-0.08=0.08$

Therefore bullet traveled 0.08 m that is 8 cm

Therefore the bullet will travel 8-6 = 2 cm further  

Answer 2

Explanation:

The bullet will cover a distance of 8 cm inside the wooden block before coming to rest.

Therefore it will travel 2 cm further

Explanation:

The bullet enters with velocity m/s after travelling a distance of 0.06 m (6 cm) its velocity becomes m/s.

now we can clearly see that

initial velocity   m/s

final velocity m/s

distance traveled m

Now using 3rd equation of motion

clearly it is seen that the velocity is decreasing

now let us calculate distance traveled by the bullet before it comes to rest

Therefore bullet traveled 0.08 m that is 8 cm

Therefore the bullet will travel 8-6 = 2 cm further  

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