Solved Problem on Collisions

A bullet of mass 15 g collides with a wooden block of mass 2.985 kg, suspended horizontally by two wires, and the bullet lodges in the block, and the whole system rises 5 cm relative to the initial position, assuming that the wires remain parallel. This system is known as a ballistic pendulum. Determine:
a) The speed of the bullet when it hits the block;
b) The speed acquired by the bullet-block system;
c) The energy lost in the collision.

Problem data:

Bullet mass: m = 15 g;
Block mass: M = 2.985 kg;
Height that the system rises after the collision: h = 5 cm;
Acceleration due to gravity: g = 9.8 m/s².

Solution

First, we convert the mass of the bullet, given in grams (g), to kilograms (kg), and the height at which the system rises, is given in centimeters (cm), to meters (m) used in the International System of Units (SI)

\[ \begin{gather} m=15\;\cancel{\mathrm{g}}\times\frac{1.10^{-3}\;\mathrm{kg}}{1\;\cancel{\mathrm{g}}}=1.5\times 10\times 10^{-3}\;\mathrm{kg}=1.5\times 10^{-2}\;\mathrm{kg}\\[10pt] h=5\;\cancel{\mathrm{cm}}\times\frac{1.10^{-2}\;\mathrm{m}}{1\;\cancel{\mathrm{cm}}}=5\times 10{-2}\;\mathrm{cm} \end{gather} \]

a) Applying the Law of Conservation of Linear Momentum and the Principle of Conservation of Mechanical Energy to the system before and after the impact of the bullet against the block (Figure 1).

Conservation of Linear Momentum:

Before the impact, the bullet of mass m has velocity v_b, and the block of mass M is at rest, v_B = 0. Immediately after the collision, the bullet-block system, of mass m+M, has velocity V. The momentum before collision \( \left( Q_{i} \right) \) is equal to the momentum after shock \( \left( Q_{f} \right) \).

Figure 1

\[ \begin{gather} Q_{i}=Q_{f}\\[5pt] mv_{b}+Mv_{B}=(m+M)V\\[5pt] mv_{b}+M\times 0=(m+M)V\\[5pt] mv_{b}=(m+M)V\\[5pt] v_{b}=\frac{m+M}{m}V \tag{I} \end{gather} \]

Conservation of Mechanical Energy:

We choose a Reference Level (R.L.) in the middle of the resting block. At the moment of impact, the bullet-block system has kinetic energy \( \left( K_{i} \right) \), the potential energy \( \left( U_{i} \right) \) is zero, and the height relative to the reference is zero (Figure 1). When the system reaches its maximum height, the system has potential energy \( \left( U_{f} \right) \), and the kinetic energy \( \left( K_{f} \right) \) is zero, the velocity of the system is equal to zero (the system stops for an instant before returning).

Figure 2

equating the mechanical energy \( \left( E_{m} \right) \) of the system at the moment of impact and at the moment it reaches the maximum height

\[ \begin{gather} E_{m i}=E_{m f}\\[5pt] K_{i}+U_{i}=K_{f}+U_{f}\\[5pt] \frac{(m+M)V^{2}}{2}+(m+M)g\times 0=\frac{(m+M)\times 0^{2}}{2}+(m+M)gh\\[5pt] \frac{\cancel{(m+M)}V^{2}}{2}=\cancel{(m+M)}gh\\[5pt] \frac{V^{2}}{2}=gh\\[5pt] V^{2}=2gh\\[5pt] V=\sqrt{2gh\;} \tag{II} \end{gather} \]

substituting equation (II) into equation (I)

\[ \begin{gather} v_{b}=\frac{m+M}{m}\sqrt{2gh\;}\\[5pt] v_{b}=\frac{0.015+2.985}{0.015}\times\sqrt{2\times 9.8\times 0,05\;}\\[5pt] v_{b}=\frac{3}{0.015}\times\sqrt{0.98\;}\\[5pt] v_{b}=\frac{3}{0.015}\times 0,99 \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {v_{b}=198\;\mathrm{m/s}} \end{gather} \]

b) From equation (II), we immediately have that

\[ \begin{gather} V=\sqrt{2\times 9.8\times 0.05\;}\\[5pt] V=\sqrt{0.98\;} \end{gather} \]

\[ \bbox[#FFCCCC,10px] {V=0.99\;\mathrm{m/s}} \]

c) Applying the Principle of Conservation of Mechanical Energy again, before the collision, the bullet has kinetic energy \( \left( K_{i b} \right) \), its potential energy \( \left( U_{i b} \right) \) is zero, and its height relative to the Reference Level is zero. The kinetic and potential energies of block \( \left( K_{i B} \;\text{and}\; U_{i B} \right) \) are zero, its velocity is equal to zero, and its height, relative to the Reference Level, is equal to zero.
During the collision part of the initial energy is dissipated \( \left( E_{\small D} \right) \), this dissipated energy must be added to the final mechanical energy of the system, and what is left makes the bullet-block system oscillate, so when the system reaches the maximum height, its kinetic energy \( \left( K_{f} \right) \) will be zero, the velocity of the system is zero, and the remaining energy will be in the form of potential energy \( \left( U_{f} \right) \).

Figure 3

Using Figure 3

\[ \begin{gather} E_{m i}=E_{m f}+E_{\small D}\\[5pt] K_{i b}+U_{i b}+K_{i\small B}+U_{i\small B}=K_{f}+U_{f}+E_{\small D}\\[5pt] \frac{mv_{b}^{2}}{2}+mgh+\frac{MV_{\small B}^{2}}{2}+Mgh=\frac{(m+M)V^{2}}{2}+(m+M)gh+E_{\small D}\\[5pt] \frac{mv_{b}^{2}}{2}+mg\times 0+\frac{M\times 0^{2}}{2}+Mg\times 0=\frac{(m+M)\times 0^{2}}{2}+(m+M)gh+E_{\small D}\\[5pt] \frac{mv_{b}^{2}}{2}=(m+M)gh+E_{\small D}\\[5pt] E_{\small D}=\frac{mv_{b}^{2}}{2}-(m+M)gh\\[5pt] E_{\small D}=\frac{0.015\times 198^{2}}{2}-(0.015+2.985)\times 9.8\times 0.05\\[5pt] E_{D}=294.03-1.47 \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {E_{D}=295.5\;\mathrm{J}} \end{gather} \]

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .