Question

An automobile engine propels a $1000kg$ car (A) along a leveled road at a speed of $36km/h$. Find the power of the opposing frictional force of$100N$. Now, suppose after traveling a distance $200m$, this car collides with another stationary car (B) of the same mass and comes to rest. Let its engine also stop at the same time. Now the car (B) starts moving on the same level road without getting its engine started. Find the speed of the car (B) just after the collision.

Answer 1

Step 1: Given data

The mass of the two cars are $m_A=m_B=1000kg.$

The speed of car A is $v_A=36km/h=\frac{36\times 1000}{3600}=10m/s$ and that of car B is $v_B=0.$

The frictional force on the car is $f=100N.$

The distance traveled by car A is $s=200m.$

The velocity of car A after the collision is $v_A\text{'}=0$ and car B is $v_B\text{'}$

Step 2: Formulas used

Power is the net work done by a body in a unit of time.

The power of a dynamic body moving with a velocity $v$ is defined by the form as,
$P=F·v$
where, $F$ is the applied force on the body.

From Newton's second law of motion,
$F=ma$
where $F$ is the force on the body, $m$ is the mass of the body and $a$ is the acceleration

From the equations of motion,
$$\begin{gathered} v=u+at \\ s=ut+\frac{1}{2}a{t}^2 \end{gathered}$$
where $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration, $t$ is the time taken and $s$ is the distance traveled.

The linear momentum of a body is the product of mass and its velocity, i.e, $L=mv$.
If two bodies of mass $m_A$ and $m_B$ moving with velocities $v_A$ and $v_B$ collide.
After the collision, the velocities of the two bodies are $v_A\text{'}$ and $v_B\text{'}$, then from conservation of momentum,
From the conservation of momentum,
$m_A{v}_A+m_B{v}_B=m_A{v}_A\text{'}+m_B{v}_B\text{'}$.

Step 3: Find power of opposing frictional force

Substituting in the formula for power,
$\begin{array}{rcl}P& =& F·v\\ & =& 100\times 10\\ \therefore P& =& 1000W.\end{array}$

Therefore, the power of frictional force is $1000W$.

Step 4: Find velocity of car at the time of collision

The acceleration caused by the frictional force can be calculated from Newton's second law of motion. So the acceleration is,
$\begin{array}{cc}& 100=1000a\\ \Rightarrow & a=0.1m/s^2\end{array}$

But the acceleration by frictional force is the direction opposite to that of motion. So it is essentially deceleration.
Hence, $a=-0.1m/s^2$ (negative because it is decelration)

Here, $s=200$

Substituting in the second equation of motion,
$$\begin{gathered} \begin{array}{ccc}& 200=10\times t+\frac{1}{2}\left(-0.1\right)t^2& \\ \Rightarrow & 0.05t^2-10t+200=0& \\ \Rightarrow & t=\frac{10\pm \sqrt{{10}^2-4\times 0.05\times 200}}{2\times 0.05}& \left[\because a{x}^2+bx+c=0\Rightarrow x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\right]\\ \Rightarrow & t=\frac{10\pm \sqrt{100-40}}{0.1}& \\ \Rightarrow & t=10\left(10\pm \sqrt{60}\right)s& \end{array} \\ \begin{array}{ccc}\Rightarrow & t=10\left(10+\sqrt{60}\right)s& t=10\left(10-\sqrt{60}\right)s\end{array} \end{gathered}$$

In our case, we will take $t=10\left(10-\sqrt{60}\right)$ because the other value describes the situation where the car's deceleration (negative acceleration) makes its velocity negative thereby making the car turn around.

So, the velocity of the car at $t=10\left(10-\sqrt{60}\right)$, when it will have travelled $200m$ is,
$\begin{array}{rcl}v& =& 10+\left(-0.1\right)\left[10\left(10-\sqrt{60}\right)\right]\\ & =& 10-10+\sqrt{60}\\ & =& \sqrt{60}\\ & =& 7.75m/s\end{array}$

Step 5: Finding the velocity of car B after the collision

From the conservation of momentum,

$\begin{array}{cc}& m_A{v}_A+m_B{v}_B=m_A{v}_A\text{'}+m_B{v}_B\text{'}\\ \Rightarrow & 1000·7.75+1000·0=1000·0+1000v_B\text{'}\\ \Rightarrow & v_B\text{'}=\frac{1000·7.75}{1000}\\ \therefore & v_B\text{'}=7.75m/s\end{array}$

Therefore, the velocity of the car B after the collision is $7.75m/s$.

Hence,

1. The power of the frictional force is $1000W$.
2. The velocity of the car B after the collision is $7.75m/s$.