A bullet-block collision is a two-stage mechanics problem. During the very short impact, horizontal momentum is
approximately conserved. After the collision, the moving block or block-bullet system may keep moving on a frictionless
surface or lose energy through kinetic friction.
Stage 1: momentum during the collision
The collision happens so quickly that external horizontal forces usually deliver a negligible impulse. Therefore,
horizontal momentum is conserved during the impact.
Bullet embeds in the block.
\[
\begin{aligned}
m_bv_{b,i} &= (m_b+M)v
\end{aligned}
\]
Here \(m_b\) is the bullet mass, \(M\) is the block mass, \(v_{b,i}\) is the bullet speed before impact,
and \(v\) is the common speed immediately after impact.
Bullet passes through the block.
\[
\begin{aligned}
m_bv_{b,i} &= m_bv_{b,f}+Mv
\end{aligned}
\]
In this case, \(v_{b,f}\) is the bullet exit speed and \(v\) is the block speed immediately after the bullet leaves it.
Finding the post-collision speed
If the bullet embeds, the common speed is:
\[
\begin{aligned}
v &= \frac{m_bv_{b,i}}{m_b+M}
\end{aligned}
\]
If the bullet passes through:
\[
\begin{aligned}
v &= \frac{m_b(v_{b,i}-v_{b,f})}{M}
\end{aligned}
\]
Finding the bullet speed
If the post-collision block speed is known, the original bullet speed can be found by rearranging momentum conservation.
Embedded case.
\[
\begin{aligned}
v_{b,i} &= \frac{m_b+M}{m_b}v
\end{aligned}
\]
Pass-through case.
\[
\begin{aligned}
v_{b,i} &= v_{b,f}+\frac{M}{m_b}v
\end{aligned}
\]
Stage 2: sliding with kinetic friction
If the block or block-bullet system slides on a rough horizontal surface and eventually stops, kinetic friction does
negative work. The friction force is:
\[
\begin{aligned}
f_k &= \mu_k m_{\mathrm{slide}}g
\end{aligned}
\]
If the sliding object stops after distance \(d\), the work done by friction equals the kinetic energy after impact:
Work-energy relation for stopping distance.
\[
\begin{aligned}
\frac12m_{\mathrm{slide}}v^2 &= \mu_k m_{\mathrm{slide}}gd
\end{aligned}
\]
The sliding mass cancels, giving:
\[
\begin{aligned}
v &= \sqrt{2\mu_k gd}
\end{aligned}
\]
Or, if \(v\) is known and the distance is unknown:
\[
\begin{aligned}
d &= \frac{v^2}{2\mu_k g}
\end{aligned}
\]
Which mass slides?
The sliding mass depends on the impact model:
-
If the bullet embeds, the sliding mass is \(m_{\mathrm{slide}}=m_b+M\).
-
If the bullet passes through, the block is the object whose stopping distance is usually measured, so
\(m_{\mathrm{slide}}=M\).
Although the sliding mass appears in the friction work equation, it cancels when solving \(v=\sqrt{2\mu_kgd}\).
Kinetic energy is not conserved during the impact
Bullet-block impacts are inelastic. Momentum can be conserved while kinetic energy decreases. For an embedding collision:
\[
\begin{aligned}
K_i &= \frac12m_bv_{b,i}^2\\
K_{\mathrm{after}} &= \frac12(m_b+M)v^2
\end{aligned}
\]
The kinetic energy lost in the impact is:
\[
\begin{aligned}
K_{\mathrm{lost,impact}} &= K_i-K_{\mathrm{after}}
\end{aligned}
\]
For a pass-through collision:
\[
\begin{aligned}
K_{\mathrm{after}}
&=
\frac12m_bv_{b,f}^2+\frac12Mv^2
\end{aligned}
\]
The missing kinetic energy becomes internal energy, heat, sound, and deformation.
Energy lost by friction after collision
If friction brings the sliding object to rest, friction removes:
\[
\begin{aligned}
W_{\mathrm{friction, lost}}
&=
\mu_km_{\mathrm{slide}}gd
\end{aligned}
\]
This is not collision energy loss. It happens after the collision while the block or combined system slides.
Worked example: bullet embeds and slides
Suppose a \(0.015\ \mathrm{kg}\) bullet moving at \(420\ \mathrm{m\,s^{-1}}\) embeds in a
\(2.00\ \mathrm{kg}\) block. The surface has \(\mu_k=0.30\).
Find common speed after impact.
\[
\begin{aligned}
v
&=
\frac{m_bv_{b,i}}{m_b+M}\\
&=
\frac{(0.015)(420)}{0.015+2.00}\\
&\approx 3.13\ \mathrm{m\,s^{-1}}
\end{aligned}
\]
Find sliding distance until rest.
\[
\begin{aligned}
d
&=
\frac{v^2}{2\mu_k g}\\
&=
\frac{(3.13)^2}{2(0.30)(9.80665)}\\
&\approx 1.66\ \mathrm{m}
\end{aligned}
\]
Energy loss during impact.
\[
\begin{aligned}
K_i &= \frac12(0.015)(420)^2\approx1323\ \mathrm{J}\\
K_{\mathrm{after}} &= \frac12(2.015)(3.13)^2\approx9.84\ \mathrm{J}\\
K_{\mathrm{lost,impact}} &\approx1313\ \mathrm{J}
\end{aligned}
\]
Most of the bullet’s initial kinetic energy is lost during the inelastic impact.
Formula summary
| Case |
Formula |
Meaning |
| Bullet embeds |
\(m_bv_{b,i}=(m_b+M)v\) |
Momentum conservation for a perfectly inelastic impact |
| Bullet passes through |
\(m_bv_{b,i}=m_bv_{b,f}+Mv\) |
Momentum conservation when the bullet exits |
| Embedded speed |
\(v=\frac{m_bv_{b,i}}{m_b+M}\) |
Common speed after embedding |
| Pass-through block speed |
\(v=\frac{m_b(v_{b,i}-v_{b,f})}{M}\) |
Block speed after the bullet exits |
| Friction stopping speed |
\(v=\sqrt{2\mu_kgd}\) |
Post-collision speed from stopping distance |
| Sliding distance |
\(d=\frac{v^2}{2\mu_kg}\) |
Distance needed for kinetic friction to stop the sliding object |
| Impact energy loss |
\(K_{\mathrm{lost,impact}}=K_i-K_{\mathrm{after}}\) |
Kinetic energy transformed during the collision |
| Friction energy loss |
\(W_{\mathrm{friction,lost}}=\mu_km_{\mathrm{slide}}gd\) |
Energy transformed by friction after impact |
Use momentum for the impact and work-energy for the sliding motion. Do not conserve kinetic energy during the bullet-block collision.
