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P 31

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Found in: Page 247

### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271

# Question: A ${\mathbf{12}}{\mathbf{.}}{\mathbf{0}}{\mathbf{-}}{\mathbit{g}}$ wad of sticky clay is hurled horizontally at a 100-g wooden block initially at rest on a horizontal surface. The clay sticks to the block. After impact, the block slides before 7.50 m coming to rest. If the coefficient of friction between the block and the surface is 0.650, what was the speed of the clay immediately before impact?

The speed of the clay immediately before impact is, ${v}_{C}=91.2\text{m/s}$.

See the step by step solution

## Step 1: Concept

Step 2: Stating given data

$P=mv$ (1)

Here, P represents the momentum, m represents the mass of the particle, v represents the speed of the particle.

## Step 2: Stating given data

Mass of the clay, ${m}_{C}=12\text{g}=0.0120\text{kg}$.

Mass of the block, ${m}_{B}=100\text{g}=\text{0.100}\text{kg}$.

The coefficient of friction between the block and the surface is, $\mu =0.650$.

The distance traveled by block before coming to rest, $d=7.50\text{m}$.

## Step 3: Calculating speed of clay immediately before the impact

The collision between the clay and the wooden block is not elastic. In this collision which is known as inelastic collision, the momentum will remain conserved.

Thus, if the total momentum before the impact and total momentum after the momentum are the same, then we can write

${P}_{Bi}+{P}_{Ci}={P}_{Bf}+{P}_{Cf}$

From equation (1), we can write the above expression as

${m}_{B}{v}_{Bi}+{m}_{C}{v}_{Ci}={m}_{B}{v}_{Bf}+{m}_{C}{v}_{Cf}$

Here i and f subscript represent the initial and final values.

The block is initially at rest so that the velocity will be zero. And the clay is moving with a velocity ${v}_{C}$ before the impact. After the impact, they stick together and move with a velocity of ${v}_{CB}$.

The above expression can be written as;

$\begin{array}{rcl}{m}_{B}0+{m}_{C}{v}_{C}& =& {m}_{B}{v}_{CB}+{m}_{C}{v}_{CB}\\ {m}_{C}{v}_{C}& =& \left({m}_{B}+{m}_{C}\right){v}_{CB}\\ {v}_{C}& =& \frac{\left({m}_{B}+{m}_{C}\right)}{{m}_{C}}{v}_{CB}\end{array}$

From the conservation of energy, the energy equation in the presence of friction forces can be written as

$\Delta K=W$ (3)

$\Delta K$ is the kinetic energy of the clay block system after the impact that can be written as

$\Delta K=\frac{1}{2}\left({m}_{B}+{m}_{C}\right){v}_{CB}^{2}$

W is the work done that can be calculated as

$W={F}_{f}d$

Here ${F}_{f}$ is the frictional force, that is ${F}_{f}=\mu \left({m}_{B}+{m}_{c}\right)g$.

Now, substituting the values in equation 3, we get

$\begin{array}{rcl}\frac{1}{2}\left({m}_{B}+{m}_{c}\right){v}_{CB}^{2}& =& \mu \left({m}_{B}+{m}_{c}\right)g\\ {v}_{CB}& =& \sqrt{2\mu gd}\end{array}$ (4)

From equations 2 and 4, we can write

${v}_{C}=\frac{\left({m}_{B}+{m}_{C}\right)}{{m}_{C}}\sqrt{2\mu gd}$

Substituting the values in the above expression, we get

${v}_{C}=\frac{\left(0.100\text{kg}+0.0120\text{kg}\right)}{0.0120\text{kg}}\sqrt{2\left(0.650\right)\left(9.80{\text{m/s}}^{\text{2}}\right)\left(7.50\text{m}\right)}\phantom{\rule{0ex}{0ex}}=9.33×9.77\text{m/s}\phantom{\rule{0ex}{0ex}}{v}_{C}=91.20\text{m/s}$

Thus, the speed of the clay immediately before impact is ${v}_{C}=91.2\text{m/s}$