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Q11OQ

Expert-verifiedFound in: Page 281

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**A ${\mathbf{10}}{\mathbf{.0}}{\mathbf{}}{\mathbf{-}}{\mathit{g}}$bullet is fired into a $\mathbf{200}\mathbf{}\mathbf{-}\mathit{g}$block of wood at rest on a horizontal surface. After impact, the block slides${\mathbf{8}}{\mathbf{.0}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathit{m}}$ before coming to rest. If the coefficient of friction between the block and the surface is,${\mathbf{0}}{\mathbf{.400}}$ what is the speed of the bullet before impact?**

** (a) ${\mathbf{106}}{\mathbf{}}{\mathbf{}}{\mathit{m}}{\mathbf{/}}{\mathit{s}}$**

** (b)${\mathbf{166}}{\mathbf{}}{\mathbf{}}{\mathit{m}}{\mathbf{/}}{\mathit{s}}$**

**(c)${\mathbf{226}}{\mathbf{}}{\mathbf{}}{\mathit{m}}{\mathbf{/}}{\mathit{s}}$**

**(d)${\mathbf{286}}{\mathbf{}}{\mathbf{}}{\mathit{m}}{\mathbf{/}}{\mathit{s}}$**

**(e) none of those answers is correct**

The speed of the bullet before impact is, ${v}_{b}=166\text{\hspace{0.17em}m/s}$makingchoice (b) the correct answer.

**Let us assume that the masses and initial velocities of both particles are known. Then, we can solve the below equation to calculate the final velocities in terms of the initial velocities because there are two equations and two unknowns:**

$\begin{array}{l}{v}_{1f}=\left(\frac{{m}_{1}-{m}_{2}}{{m}_{1}+{m}_{2}}\right){v}_{1i}+\left(\frac{2{m}_{2}}{{m}_{1}+{m}_{2}}\right){v}_{2i}\\ {v}_{2f}=\left(\frac{2{m}_{1}}{{m}_{1}+{m}_{2}}\right){v}_{1i}+\left(\frac{{m}_{1}-{m}_{2}}{{m}_{1}+{m}_{2}}\right){v}_{2i}\end{array}$

Here,

${m}_{1}=$Mass of the first particle.

${m}_{2}=$Mass of the second particle

${v}_{1i}=$ Initial velocity of the first particle

${v}_{2i}=$Initial velocity of the second particle

${v}_{1f}=$Final velocity of the first particle

${v}_{2f}=$Final velocity of the second particle

** **We find the velocity of mass-1 using the equation derived for an elastic collision.

Before the collision, the bullet with mass${m}_{1}=10.0\text{\hspace{0.17em}g}$has speed, ${v}_{1i}={v}_{b}$and the block with mass ${m}_{2}=200\text{\hspace{0.17em}g}$has speed ${v}_{2i}=0$

After the collision, the objects have a common speed (velocity),${v}_{1f}={v}_{2f}=v$.

The collision of the bullet with the block is completely inelastic.

From the conservation of momentum theorem, we can write the momentum equation as follows:

${m}_{1}{v}_{1i}+{m}_{2}{v}_{2}={m}_{1}{v}_{1f}+{m}_{2}{v}_{2f}$

Put the values in the above equation.

$\begin{array}{c}{m}_{1}{v}_{b}=({m}_{1}+{m}_{2})v\\ {v}_{b}=v\left(\frac{{m}_{1}+{m}_{2}}{{m}_{1}}\right)\end{array}$ (1)

The kinetic friction can be written as follows:

${f}_{k}={\mu}_{k}N$

Here,is the coefficient of friction, and N is the normal/contact force.

It slows down the block with anacceleration of magnitude $a={\mu}_{k}g$.

The block slides to a stop through a distance. $d=800\text{\hspace{0.17em}m}$

Using the law of motion equation ${v}_{f}^{2}={v}_{i}^{2}+2ad$, we find the speed of the block after the collision.

$v=\sqrt{2{\mu}_{k}gd}$

Put the values in the above equation.

$\begin{array}{c}v=\sqrt{2\left(0.400\right)\left(9.8{\text{\hspace{0.17em}m/s}}^{2}\right)\left(8.00\text{\hspace{0.17em}m}\right)}\\ =7.92\text{\hspace{0.17em}m/s}\end{array}$

Put the values in the above equation.

$\begin{array}{c}{v}_{b}=(7.92\text{\hspace{0.17em}m/s})\left(\frac{10+200}{10.0}\right)\\ =166\text{\hspace{0.17em}m/s}\end{array}$

Thus, the speed of the bullet before impact is, ${v}_{b}=166\text{\hspace{0.17em}m/s}$making choice (b) the correct answer.

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