Chapter 13.3, Problem 13.149P

Bullet B weighs 0.5 oz and blocks A and C both weigh 3 lb. The coefficient of friction between the blocks and the plane is μ_k = 0.25. Initially, the bullet is moving at v₀ and blocks A and C are at rest (Fig. 1). After the bullet passes through A, it becomes embedded in block C and all three objects come to stop in the positions shown (Fig. 2). Determine the initial speed of the bullet v₀.

Fig. P13.149

To determine

Find the initial speed $\left(v_0\right)$ of the bullet.

Answer to Problem 13.149P

The initial speed $\left(v_0\right)$ of the bullet is $\underset{\_}{497.08\text{ft}/\text{s}}$.

Explanation of Solution

Given information:

The weight of the bullet B $\left(W_B\right)$ is $5\text{oz}$.

The weight of the block A $\left(W_A\right)$ is $3\text{lb}$.

The weight of the block C $\left(W_C\right)$ is $3\text{lb}$.

The coefficient of friction between the blocks and plane $\left({\mu }_k\right)$ is 0.25.

The distance between the upper block A and lower block A $\left(d_A\right)$ is $6\text{in}\text{.}$.

The distance between the upper block C and lower block C $\left(d_C\right)$ is $\text{4in}\text{.}$.

The acceleration due to gravity (g) is $32.2\text{ft}/{\text{s}}^{\text{2}}$.

Calculation:

Show the free body diagram of the block A as the bullet passes through it as Figure (1).

The expression for the principle of conservation of momentum to the bullet and block A as follows;

$m_A{\left(v_A\right)}_0+m_B{v}_0=m_A{v}_A+m_B{v}_1$

Here, $m_A$ is the mass of the block A, ${\left(v_A\right)}_0$ is the initial velocity of the block A, $m_B$ is the mass of the bullet B, $v_0$ is the initial velocity of the bullet before passing through A, $v_A$ is the velocity of the block A just after the bullet passes through and $v_1$ is the velocity of the bullet after it passes through block A.

Since the block A is at rest initially, so the velocity ${\left(v_A\right)}_0$ will be zero.

Substitute 0 for ${\left(v_A\right)}_0$.

$\begin{array}{l}0+m_B{v}_0=m_A{v}_A+m_B{v}_1\\ v_0=v_1+\frac{m_A{v}_A}{m_B}\end{array}$ (1)

Calculate the mass of the bullet $\left(m_B\right)$ using the formula:

$m_B=\frac{W_B}{g}$

Substitute $5\text{oz}$ for $W_B$ and $32.2\text{ft}/{\text{s}}^2$ for g.

$\begin{array}{c}{m}_B=\frac{5\text{oz}}{32.2\text{ft}/{\text{s}}^2}\\ =\frac{5}{32.2}\times \left(\frac{1\text{lb}}{16\text{oz}}\right)\\ =970.5\times {10}^{-6}\text{lb}\text{.}{\text{s}}^2/\text{ft}\end{array}$

Calculate the mass of the block A $\left(m_A\right)$ using the formula:

$m_A=\frac{W_A}{g}$

Substitute $3\text{lb}$ for $W_A$ and $32.2\text{ft}/{\text{s}}^2$ for g.

$\begin{array}{c}{m}_A=\frac{3\text{lb}}{32.2\text{ft}/{\text{s}}^2}\\ =93.168\times {10}^{-3}\text{lb}\text{.}{\text{s}}^2/\text{ft}\end{array}$

The expression for the normal force acting on the block A as follows;

$N_A=m_{A}g$

The expression for the work done $\left[{\left(U_{1-2}\right)}_A\right]$ by frictional force acting on the block A as it slides through the distance $d_A$ as follows;

${\left(U_{1-2}\right)}_A=-{\mu }_k{N}_A{d}_A$

Substitute $m_{A}g$ for $N_A$.

${\left(U_{1-2}\right)}_A=-{\mu }_k{m}_{A}g{d}_A$

The expression for the initial kinetic energy of the block A ${\left(T_1\right)}_A$ while the bullet just passes through the block A as follows;

${\left(T_1\right)}_A=\frac{1}{2}{m}_A{v}_A^2$

The final kinetic energy of the block A ${\left(T_2\right)}_A$ after the bullet leaves the block is zero as the block A comes to rest finally.

The expression for the principle of work-energy to the block A after the bullet just passes through it as follows;

${\left(T_1\right)}_A+{\left(U_{1-2}\right)}_A={\left(T_2\right)}_A$

Substitute $\frac{1}{2}{m}_A{v}_A^2$ for ${\left(T_1\right)}_A$, ${\mu }_k{m}_{A}g{d}_A$ for ${\left(U_{1-2}\right)}_A$, and 0 for ${\left(T_2\right)}_A$.

$\begin{array}{l}\frac{1}{2}{m}_A{v}_A^2-{\mu }_k{m}_{A}g{d}_A=0\\ \frac{1}{2}{m}_A{v}_A^2={\mu }_k{m}_{A}g{d}_A\\ v_A^2=\frac{2{\mu }_k{m}_{A}g{d}_A}{m_A}\\ v_A=\sqrt{2{\mu }_{k}g{d}_A}\end{array}$

Substitute 0.25 for ${\mu }_k$, $32.2\text{ft}/{\text{s}}^2$ for g, and $6\text{in}\text{.}$ for $d_A$.

$\begin{array}{c}{v}_A=\sqrt{2\left(0.25\right)\left(32.2\text{ft}/{\text{s}}^2\right)\left(6\text{in}\right)}\\ =\sqrt{\left(16.1\right)\left(6\times \frac{0.08333\text{ft}}{1\text{in}}\right)}\\ =\sqrt{\left(16.1\right)\left(0.5\right)}\\ =2.8372\text{}\text{ft}/\text{s}\end{array}$

Show the free body diagram of the block C as the bullet passes through it as in Figure (2).

The expression for the principle of conservation of momentum to the bullet and block A as follows;

$m_C{\left(v_C\right)}_0+m_B{v}_1=m_C{v}_C+m_B{v}_C$

Here, $m_C$ is the mass of the block C, ${\left(v_C\right)}_0$ is the initial velocity of the block C and $v_C$ is the velocity of the block and bullet after the bullet gets embedded in the block C.

The initial velocity of the block C ${\left(v_C\right)}_0$ is zero as the block is at rest initially.

Substitute 0 for ${\left(v_C\right)}_0$.

$\begin{array}{l}0+m_B{v}_1=m_C{v}_C+m_B{v}_C\\ v_1=\frac{\left(m_B+m_C\right)v_C}{m_B}\end{array}$ (2)

Calculate the mass of the block C $\left(m_C\right)$ using the formula:

$m_C=\frac{W_C}{g}$

Substitute $3\text{lb}$ for $W_C$ and $32.2\text{ft}/{\text{s}}^2$ for g.

$\begin{array}{c}{m}_C=\frac{3\text{lb}}{32.2\text{ft}/{\text{s}}^2}\\ =93.168\times {10}^{-3}\text{lb}\text{.}{\text{s}}^2/\text{ft}\end{array}$

The expression for the normal force acting on the block C as follows:

$N_C=m_{C}g$

The expression for the work done $\left[{\left(U_{1-2}\right)}_C\right]$ by frictional force acting on the block C as it slides through the distance $d_C$ as follows:

$\begin{array}{c}{\left(U_{1-2}\right)}_C=-{\mu }_k{N}_C{d}_C\\ =-{\mu }_k{m}_{C}g{d}_C\end{array}$

The expression for the initial kinetic energy of the block C ${\left(T_1\right)}_C$ while the bullet just gets embedded in the block C as follows:

${\left(T_1\right)}_C=\frac{1}{2}\left(m_B+m_C\right)v_C^2$

The final kinetic energy of the block C with bullet embedded ${\left(T_2\right)}_C$ is zero as the block C comes to rest finally.

The expression for the principle of work-energy to the block C after the bullet just gets embedded in the block as follows:

${\left(T_1\right)}_C+{\left(U_{1-2}\right)}_C={\left(T_2\right)}_C$

Substitute $\frac{1}{2}\left(m_B+m_C\right)v_C^2$ for ${\left(T_1\right)}_C$, $-{\mu }_k{m}_{C}g{d}_C$ for ${\left(U_{1-2}\right)}_C$, and 0 for ${\left(T_2\right)}_C$.

$\begin{array}{l}\frac{1}{2}\left(m_B+m_C\right)v_C^2-{\mu }_k{m}_{C}g{d}_C=0\\ \frac{1}{2}\left(m_B+m_C\right)v_C^2={\mu }_k{m}_{C}g{d}_C\\ v_C^2=\frac{2{\mu }_k{m}_{C}g{d}_C}{m_B+m_C}\\ v_C=\sqrt{\frac{2{\mu }_k{m}_{C}g{d}_C}{m_B+m_C}}\end{array}$

Substitute 0.25 for ${\mu }_k$, $32.2\text{ft}/{\text{s}}^2$ for $g$, $4\text{in}\text{.}$ for $d_C$, $93.168\times {10}^{-3}\text{lb}\text{.}{\text{s}}^2/\text{ft}$ for $m_C$, and $970.5\times {10}^{-6}\text{lb}\text{.}{\text{s}}^2/\text{ft}$ for $m_B$.

$\begin{array}{c}{v}_C=\sqrt{\frac{2\left(0.25\right)\left(93.168\times {10}^{-3}\text{lb}\text{.}{\text{s}}^2/\text{ft}\right)\left(32.2\text{ft}/{\text{s}}^2\right)\left(4\text{in}\right)}{\left(93.168\times {10}^{-3}\text{lb}\text{.}{\text{s}}^2/\text{ft}+970.5\times {10}^{-6}\text{lb}\text{.}{\text{s}}^2/\text{ft}\right)}}\\ =\sqrt{\frac{1,500.0048\times {10}^{-3}\left(4\text{in}\right)}{93.168\times {10}^{-3}}}\\ =\sqrt{16.1\left(4\times \frac{0.08333\text{ft}}{1\text{in}}\right)}\\ =\sqrt{5.3665}\\ =2.3166\text{ft}/\text{s}\end{array}$

Substitute $2.3166\text{ft}/\text{s}$ for $v_C$, $93.168\times {10}^{-3}\text{lb}\text{.}{\text{s}}^2/\text{ft}$ for $m_C$, and $v_A$ $970.5\times {10}^{-6}\text{lb}\text{.}{\text{s}}^2/\text{ft}$ for $m_B$ in the Equation (2).

$\begin{array}{c}{v}_1=\frac{\left(m_B+m_C\right)v_C}{m_B}\\ =\frac{\left(93.168\times {10}^{-3}\text{lb}\text{.}{\text{s}}^2/\text{ft}+970.5\times {10}^{-6}\text{lb}\text{.}{\text{s}}^2/\text{ft}\right)\left(2.3166\text{ft}/\text{s}\right)}{970.5\times {10}^{-6}\text{lb}\text{.}{\text{s}}^2/\text{ft}}\\ =\frac{\left(93.168\times {10}^{-3}\right)\left(2.3166\right)}{970.5\times {10}^{-6}}\\ =\frac{215.833\times {10}^{-3}}{970.5\times {10}^{-6}}\\ =224.71\text{ft}/\text{s}\end{array}$

Calculate the initial speed of the bullet $\left(v_0\right)$ using the relation:

Consider the equation (1).

Substitute $224.71\text{ft}/\text{s}$ for $v_1$, $970.5\times {10}^{-6}\text{lb}\text{.}{\text{s}}^2/\text{ft}$ for $m_B$, $2.3166\text{}\text{ft}/\text{s}$ for $v_A$, and $93.168\times {10}^{-3}\text{lb}\text{.}{\text{s}}^2/\text{ft}$ for.$2.8372\text{ft}/\text{s}$ $m_A$ in the Equation (1).

$\begin{array}{c}{v}_0=v_1+\frac{m_A{v}_A}{m_B}\\ =224.71\text{ft}/\text{s}+\frac{\left(93.168\times {10}^{-3}\text{lb}\text{.}{\text{s}}^2/\text{ft}\right)\left(2.8372\text{}\text{ft}/\text{s}\right)}{970.5\times {10}^{-6}\text{lb}\text{.}{\text{s}}^2/\text{ft}}\\ =224.71+272.3712\\ =497.08\text{ft}/\text{s}\end{array}$

Therefore, the initial speed $\left(v_0\right)$ of the bullet is $\underset{\_}{497.08\text{ft}/\text{s}}$.