Question

Two carts on horizontal straight rails are pushed apart by an explosion of a point charge $Q$ placed between the carts. Suppose the coefficient of friction of the two are equal and $200\text{kg}$ cart travels a distance of $36\text{m}$ and stops. The distance covered by the cart weighing $300\text{kg}$ is

• A. $12\text{m}$
• B. $16\text{m}$
• C. $24\text{m}$
• D. $32\text{m}$

Answer 1

Answer: (C)

Step 1: Given that:

Two carts let A and B are pushed apart.

The coefficient of friction of both the carts are equal.

Mass of cart A ($m_A$) = $200kg$

Mass of cart b($m_B$) = $300kg$

The distance moved by cart A that is $s_A$ = $36m$

Step 2: Formula and concept used:

The distance moved by a body of mass m on a surface with the friction coefficient $\mu$ is given as;

$s=\frac{p^2}{2\mu m^{2}g}$

Where p is the momentum in the body and g is the acceleration due to gravity.

At constant $\mu$ for two a body

$s\propto \frac{1}{m^2}$

Step 3: Calculation of the distance covered by the cart weighing 300kg:

Therefore using the relation $s\propto \frac{1}{m^2}$ ,

(Note: Here we are considering that the momentum of both the bodies are equal.)

Now, we have;

For cart A;

$s_A=k\frac{1}{m_A^2}$ .............(1)

And if the distance travelled by cart B is $s_B$ then

$s_B=k\frac{1}{m_B^2}$ ............(2)

From equation 1) divided by equation 2), we get;

$\frac{s_A}{s_B}=\frac{k\frac{1}{m_A^2}}{k\frac{1}{m_B^2}}$

$\frac{s_A}{s_B}=\frac{m_B^2}{m_A^2}$

Putting the given values, we get;

$\frac{36m}{s_B}=\frac{{\left(300kg\right)}^2}{{\left(200kg\right)}^2}$

$\frac{36}{s_B}=\frac{300\times 300}{200\times 200}$

$\frac{36}{s_B}=\frac{9}{4}$

$9s_B=36\times 4$

$s_B=\frac{144}{9}$

$s_B=16m$

Thus,

The cart B with a mass 300kg will move to a distance of $16m$.