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P 32

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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 wad of sticky clay of mass m is hurled horizontally at a wooden block of mass M initially at rest on a horizontal surface. The clay sticks to the block. After impact, the block slides a distance d before coming to rest. If the coefficient of friction between the block and the surface is $\mathbit{\mu }$, what was the speed of the clay immediately before impact?

The speed of the clay immediately before impact is, ${v}_{c}=\frac{\left(M+m\right)}{m}\sqrt{2\mu gd}$.

See the step by step solution

## Step 1: Concept

The linear momentum of an object can be written as

$P=mv$ (1)

Here, represents the momentum, represents the mass of the particle, represents the speed of the particle.

## Step 2: Stating given data

Mass of the clay, m.

Mass of the block, M.

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

The distance traveled by block before coming to rest, d.

## Step 3: Speed of clay immediately before the impact

In this case, the collision is inelastic, which means the total momentum is conserved.

Thus, the total momentum before the impact is equal to the total momentum after the momentum is the same; we can mathematically write this as

${P}_{bi}+{P}_{ci}={P}_{bf}+{P}_{cf}$

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

${m}_{b}{v}_{bi}+{m}_{c}{v}_{ci}={m}_{b}{v}_{bf}+{m}_{c}{v}_{cf}$ (2)

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

Before the impact:

The block is at rest, ${v}_{bi}=0$.

And the clay is moving with a velocity ${v}_{c}$, ${v}_{ci}={v}_{c}$

After the impact:

They stick together, behave as a single object, and move with a velocity of ${v}_{cb}$.

Substituting the values in equation 2, we get

$M0+m{v}_{c}=M{v}_{cb}+m{v}_{cb}\phantom{\rule{0ex}{0ex}}m{v}_{c}=\left(M+m\right){v}_{cb}\phantom{\rule{0ex}{0ex}}{v}_{c}=\frac{\left(M+m\right)}{m}{v}_{cb}$

The kinetic energy of the clay block system after the impact when they are stuck together can be written as

$\Delta K=\frac{1}{2}\left(M+m\right){v}_{cb}^{2}$

The work done on the system due to the frictional force can be calculated as

$W=Fd$

Here is the frictional force that is $F=\mu \left(M+m\right)g$.

This work done and the overall kinetic energy will be equal due to the energy conservation principle, which means

$\Delta K=W$ (4)

Now, substituting the values in equation 4, we get

$\frac{1}{2}\left(M+m\right){v}_{cb}^{2}=\mu \left(M+m\right)gd\phantom{\rule{0ex}{0ex}}{v}_{cb}=\sqrt{2\mu gd}$ (5)

From equations 3 and 5, we can write

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

Thus, the speed of the clay immediately before impact is ${v}_{c}=\frac{\left(M+m\right)}{m}\sqrt{2\mu gd}$.