Solved Problem on Collisions

A nail of mass 5 g is driven into a wall using a hammer of mass 495 g. The speed of the hammer just before striking the nail is 4 m/s, and the collision is perfectly inelastic. Determine:
a) The speed of the nail-hammer system immediately after the impact;
b) The energy dissipated in the collision;
c) Assuming that the nail penetrates the wall 0.5 cm at each strike, calculate the magnitude of the force, suppose constant, opposed by the wall to the penetration.

Problem data:

Nail mass: m = 5 g;
Hammer Mass: M = 495 g;
Hammer speed: v_H = 4 m/s.

Problem diagram:

Figure 1

We can replace the hammer and nail with two spheres with the same masses as the hammer and nail, the hammer-nail system will be equivalent to a perfectly inelastic collision between these spheres (Figure 1).

Solution

First, we convert the masses of the nail and the hammer, given in grams (g), to kilograms (kg) and the displacement of the nail, given in centimeters (cm), to meters (m) used in the International System of Units (SI).

\[ \begin{gather} m=5\;\cancel{\text{g}}\times\frac{1\;\text{kg}}{1 000\;\cancel{\text{g}}}=0.005\;\text{kg}\\[10pt] M=495\;\cancel{\text{g}}\times\frac{1\;\text{kg}}{1 000\;\cancel{\text{g}}}=0.495\;\text{kg}\\[10pt] \Delta S=0.5\;\cancel{\text{cm}}\times\frac{1\;\text{m}}{100\;\cancel{\text{cm}}}=0.005\;\text{m} \end{gather} \]

a) The linear momentum is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {\vec Q=m\vec v} \end{gather} \]

a) To find the speed of the system after the collision, we apply the Law of Conservation of Momentum (Figure 1), equating the initial and final momentum

\[ \begin{gather} Q_{i}=Q_{f}\\[5pt] Mv_{\small H}+mv_{n}=(M+m)V\\[5pt] Mv_{\small H}+m\times 0=(M+m)V\\[5pt] Mv_{\small H}=(M+m)V\\[5pt] Mv_{\small H}=(M+m)V\\[5pt] V=\frac{M}{(M+m)}v_{\small H}\\[5pt] V=\frac{0.495}{(0.495+0.005)}\times 4\\[5pt] V=\frac{1.98}{(0.5)} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {V=3.96\;\mathrm{m/s}} \end{gather} \]

b) The initial energy of the system is not conserved during the collision part of the energy is dissipated, after the collision, we must add the dissipated energy to the mechanical energy of the system. We choose a Reference Level (R.L.) on the motion line (Figure 2). There is no height difference during the collision, so no potential energy is involved, only kinetic energy and dissipated energy contribute to the energy of the system.

Figure 2

Before collision, the hammer has kinetic energy, \( K_{i\small H} \), and the kinetic energy of the nail, \( K_{in} \), is zero, and at the moment of collision, part of the mechanical energy is dissipated, \( E_{\small D} \), and after the collision, the hammer-nail system has kinetic energy, \( K_{f} \).
The kinetic energy is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {K=\frac{m v^{2}}{2}} \end{gather} \]

\[ \begin{gather} E_{\small H i}=E_{\small H f}+E_{\small D}\\[5pt] K_{i \small H}+K_{in}=K_{f}+E_{\small D}\\[5pt] \frac{Mv_{\small H}^{2}}{2}+\frac{mv_{n}^{2}}{2}=\frac{(M+m)V^{2}}{2}+E_{\small D}\\[5pt] \frac{Mv_{\small H}^{2}}{2}+\frac{m\times 0^{2}}{2}=\frac{(M+m)V^{2}}{2}+E_{\small D}\\[5pt] \frac{Mv_{\small H}^{2}}{2}=\frac{(M+m)V^{2}}{2}+E_{\small D}\\[5pt] E_{\small D}=\frac{Mv_{\small H}^{2}}{2}-\frac{(M+m)V^{2}}{2}\\[5pt] E_{\small D}=\frac{0.495\times 4^{2}}{2}-\frac{(0.495+0.005)\times 3.96^{2}}{2}\\[5pt] E_{\small D}=\frac{0.495\times 16}{2}-\frac{0.5\times 15.68}{2}\\[5pt] E_{\small D}=\frac{7.92}{2}-\frac{7.84}{2}\\[5pt] E_{\small D}=3.96-392 \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {E_{\small D}=0.04\;\mathrm{J}} \end{gather} \]

c) The force that the wall exerts against the hammer-nail system is found by applying Newton's Second Law

\[ \begin{gather} \bbox[#99CCFF,10px] {\vec{F}=m\vec{a}} \tag{I} \end{gather} \]

Using Newton's Third Law - Law of Action and Reaction - the force that the wall exerts on the hammer-nail system equals the force that the system exerts on the wall. The system has an initial velocity of V₀ and decelerates to a stop when the nail enters the wall, requiring another hammer strike (Figure 3). Then we must find the acceleration that the system when the nail is driven into the wall, for this, we apply the equation of velocity as a function of displacement

\[ \begin{gather} \bbox[#99CCFF,10px] {v^{2}=v_{0}^{2}+2a\Delta S} \end{gather} \]

Figure 3

\[ \begin{gather} v^{2}=v_{0}^{2}+2a\Delta S\\[5pt] a=\frac{v^{2}-v_{0}^{2}}{2\Delta S} \tag{II} \end{gather} \]

substituindo a equsubstituting equation (II) into equation (I) for the total mass of the system (M+m)

\[ \begin{gather} F=(M+m)\left(\frac{v^{2}-v_{0}^{2}}{2\Delta S}\right) \end{gather} \]

substituting the data and using the initial velocity found in part (a), v₀ = V, when the hammer hits the nail, and v = 0, when the nail stops

\[ \begin{gather} F=(0.495+0.005)\times\left(\frac{0^{2}-3.96^{2}}{2\times 0.005}\right)\\[5pt] F=-0.5\times\frac{15.68}{0.01} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {F=-784\;\mathrm{N}} \end{gather} \]

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