How do I find the energy lost after a collision when a bullet goes through a block?

Jun 2017
300
6
Lima, Peru
The problem is as follows:

The figure from below shows a projectile whose mass is $m$ goes goes through a block of mass $M=5m$ initially at rest as indicated in the figure from below. The bullet goes out from the block with a speed which is half of the speed when it entered the block. What will be the energy lost expressed in a percentage?



The alternatives given are as follows:

$\begin{array}{ll}
1.&20\,\%\\
2.&30\,\%\\
3.&50\,\%\\
4.&70\,\%\\
5.&90\,\%\\
\end{array}$

I'm confused exactly how should I understand this kind of collision when a bullet goes through a block. Is it an inelastic or an elastic collision?. I'm assuming that it is an inelastic collision as the two objects are joined at least for a brief period.

Then using this logic:

$p_1=p_2$

$m_1v_1=\left(m_1+m_2\right)v_2$

$mv_1= \left(m+5m\right)v_2$

According to this the speed of the block would be

$v_2=\frac{1}{6}v_1$

But I am given the exit speed of the bullet, then:

Initially the energy of the bullet:

$E_k=\frac{1}{2}mv^2$

The energy of the bullet after the collision:

$E_k=\frac{1}{2}m\left(\frac{1}{2}v\right)^2$

Then the subtraction would give the energy lost:

$\Delta E_k = \frac{1}{2}mv^2 - \frac{1}{2}m\left(\frac{1}{2}v\right)^2$

$\Delta E_k = \frac{1}{2}mv^2 - \frac{1}{8}mv^2$

$\Delta E_k = \frac{3}{8}mv^2$

Then the percentage lost would be:

$\frac{\Delta E_k}{E_k} = \frac{\frac{3}{8}mv^2}{\frac{1}{2}mv^2}$

$\frac{\Delta E_k}{E_k} = \frac{3}{4}=0.75$

But this results in $75\,\%$ but this is not among the alternatives. What part I missed the right path?.
 

topsquark

Math Team
May 2013
2,385
996
The Astral plane
The problem is that the collision is not perfectly inelastic so we have to do \(\displaystyle m_1 v_{1i} = m_1 v_1 + m_2 v_2\), where \(\displaystyle v_{1i} = v\), \(\displaystyle v_1 = \dfrac{1}{2} v\), and \(\displaystyle v_2\) is unknown. (At the end of a perfectly inelastic collision both objects are merged in some way. This is not the case here as the bullet exits the block.)

Also, if we take the assumption that the collision is perfectly inelastic the final kinetic energy would have been \(\displaystyle KE_f = \dfrac{1}{2} (m_1 + m_2) v_2 ^2\) because they both would have the same speed after the collision. You have to include both blocks when you calculate the final kinetic energy.

-Dan
 
Last edited:

skeeter

Math Team
Jul 2011
3,211
1,734
Texas
All collisions that do not conserve KE are inelastic. A perfectly inelastic collision is where the masses stick together and stay that way.
Maybe you should research before making assumptions.


$mv_0 = m \cdot \dfrac{v_0}{2} + 5m \cdot v_f \implies v_f = \dfrac{v_0}{10}$

$KE_0 = \dfrac{1}{2}mv_0^2$

$KE_f = \dfrac{1}{2}m\left(\dfrac{v_0}{2}\right)^2 + \dfrac{1}{2}(5m)\left(\dfrac{v_0}{10}\right)^2 = \dfrac{3}{20}mv_0^2$

$\dfrac{\frac{3}{20}mv_0^2 - \frac{1}{2}mv_0^2}{\dfrac{1}{2}mv_0^2} = \dfrac{-\frac{7}{20}}{\frac{1}{2}} = -\dfrac{7}{10}$

Energy lost is 70%
 
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