Problem on Dynamics . Help me to figure it out.

A bullet can enter
50 cm in a wood
30 cm in a wall
100 cm in water .

If there is
1) a 10 cm wood at first
2) then 2 cm wall in the second
3) Layer of water at last .

Q.) How much distance the bullet will go in the water ?
 

topsquark

Forum Staff
Apr 2008
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On the dance floor, baby!
A bullet can enter
50 cm in a wood


If there is
1) a 10 cm wood at first
These seem to contradict each other. Could you explain what you are doing in more detail?

-Dan
 
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Apr 2015
1,238
359
Somerset, England
This is a problem in simultaneous or coupled equations, that is not as bad as it seems.

So I will give you a start

Assume constant deceleration in all cases.

Let
v0 be the approach velocity of the bullet in all cases.
-a1 be the deceleration in wood
-a2 be the deceleration in the wall
-a3 be the deceleration in water

Then using \(\displaystyle {v^2} = {u^2} + 2as\) and the information about the distances which each bring the bullet to a standstill


\(\displaystyle 0 = v_0^2 - 2{a_1}*50\)

\(\displaystyle 0 = v_0^2 - 2{a_2}*30\)

\(\displaystyle 0 = v_0^2 - 2{a_3}*100\)

or

\(\displaystyle 100{a_1} = 60{a_2} = 200{a_3}\)

Which reduces 4 unknowns to 2.

If we now let
S be the unknown distance in water to be found can you generate some more equations to solve this?
 
Jan 2019
106
80
bullet’s initial kinetic energy - work done by medium = 0

for wood, $\dfrac{1}{2}mv^2 = F_a \cdot 0.5 \implies |F_a| = mv^2$

for wall, $\dfrac{1}{2}mv^2 = F_b \cdot 0.3 \implies |F_b| = \dfrac{5}{3}mv^2$

for water, $\dfrac{1}{2}mv^2 = F_c \cdot 1.0 \implies |F_c| = \dfrac{1}{2}mv^2$


$\dfrac{1}{2}mv^2 = |F_a| \cdot 0.1 + |F_b| \cdot 0.02 + |F_c| \cdot \Delta x$

solve for $\Delta x$
 
Apr 2015
1,238
359
Somerset, England
bullet’s initial kinetic energy - work done by medium = 0
You have the mass of the bullet to calculate this?
 
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