Question

A car having a mass of 10 metric tonnes rolls at $2{\text{m}}{{\text{s}}^{ - 1}}$ along a level track and collides with another car at rest having a mass of 20 metric tonnes. If the cars couple together after the collision, find the loss in kinetic energy during a collision.
A) $1{\text{KJ}}$
B) $1.33{\text{J}}$
C) $1.33{\text{KJ}}$
D) $1.33 \times {10^4}{\text{J}}$

Answer 1

Hint: Since the cars couple together after the collision, the two cars will move as a single car with a common velocity and a mass equal to the sum of the masses of the two cars. The collision is perfectly inelastic. So, the linear momentum is conserved before and after the collision. Using the linear momentum conservation theorem the common velocity of the two cars after the collision can be obtained.

Formulas used:
-The linear momentum of a body of mass $m$ and moving with velocity $v$ is given by, $p = mv$
-The kinetic energy of a body of mass $m$ and moving with velocity $v$ is given by, $K = \dfrac{1}{2}m{v^2}$

Complete step by step answer.
Step 1: List the key details mentioned in the question.
A car collides into a car at rest. After the collision, the two cars stick together and have a mass equal to the sum of the two masses of the cars and a common velocity.
The mass of the first car is ${m_1} = 10{\text{ metric tonnes}}$ and the mass of the second car is ${m_2} = 20{\text{ metric tonnes}}$ .
The initial velocity of the first car is given to be ${v_1} = 2{\text{m}}{{\text{s}}^{ - 1}}$. Since the second car is at rest initially, its velocity will be ${v_2} = 0{\text{m}}{{\text{s}}^{ - 1}}$
Let ${m_1} + {m_2} = 30{\text{ metric tonnes}}$ be the mass of the coupled cars and $v$ be the common velocity of the two cars after the collision.
Step 2: Find the common velocity of the two cars using the linear momentum conservation theorem.
Let ${p_b}$ be the linear momentum of the system before the collision and ${p_a}$ be the linear momentum of the system after the collision.
The linear momentum of a body of mass $m$ moving with velocity $v$ is given by, $p = mv$ ----- (1).
Here before the collision, the system comprises of the two cars of masses ${m_1}$ and ${m_2}$ having respective initial velocities ${v_1}$ and ${v_2}$
Using equation (1), the linear momentum of the system before the collision is
${p_b} = {m_1}{v_1} + {m_2}{v_2}$ ---------- (2).
After the collision, the system comprises two cars coupled together of mass ${m_1} + {m_2}$ moving with a common velocity $v$ .
Using equation (1), the linear momentum of the system after the collision is
${p_a} = \left( {{m_1} + {m_2}} \right)v$ ---------- (3).
The linear momentum conservation theorem demands the linear momentum of the system before and after a collision to be constant. Then according to the theorem, ${p_b} = {p_a}$ .
Thus we can equate (2) and (3) to get, ${m_1}{v_1} + {m_2}{v_2} = \left( {{m_1} + {m_2}} \right)v$ ------- (4)
Substituting the values for ${m_1} = 10{\text{ metric tonnes}}$, ${m_2} = 20{\text{ metric tonnes}}$, ${v_1} = 2{\text{m}}{{\text{s}}^{ - 1}}$ and ${v_2} = 0{\text{m}}{{\text{s}}^{ - 1}}$ in equation (4) we get, $\left[ {10 \times {{10}^3} \times 2} \right] + \left[ {\left( {20 \times {{10}^3}} \right) \times 0} \right] = \left[ {30 \times {{10}^3}} \right]v$
Simplifying we get, $20 \times {10^3} = 30 \times {10^3}v$ or, $v = \dfrac{{20 \times {{10}^3}}}{{30 \times {{10}^3}}} = \dfrac{2}{3}{\text{m}}{{\text{s}}^{ - 1}}$
Then the common velocity of the two cars is obtained as $v = \dfrac{2}{3}{\text{m}}{{\text{s}}^{ - 1}}$
Step 3: Find the loss in kinetic energy during the collision.
Let ${K_b}$ be the kinetic energy before the collision and ${K_a}$ be the kinetic energy after the collision.
Then the kinetic energy before the collision is given by, ${K_b} = \dfrac{1}{2}{m_1}{v_1}^2$ and on substituting values for ${m_1} = 10{\text{ metric tonnes}}$ and ${v_1} = 2{\text{m}}{{\text{s}}^{ - 1}}$ it will be ${K_b} = \dfrac{1}{2}10 \times {10^3} \times {2^2} = 20 \times {10^3}{\text{J}}$ or ${K_b} = 20{\text{KJ}}$
Similarly, the kinetic energy after the collision is given by, ${K_a} = \dfrac{1}{2}\left( {{m_1} + {m_2}} \right){v^2}$ and on substituting values for ${m_1} + {m_2} = 30{\text{ metric tonnes}}$ and $v = \dfrac{2}{3}{\text{m}}{{\text{s}}^{ - 1}}$ it will be ${K_a} = \dfrac{1}{2} \times 30 \times {10^3} \times {\dfrac{2}{3}^2} = \dfrac{{20 \times {{10}^3}}}{3}{\text{J}}$ or ${K_a} = \dfrac{{20}}{3}{\text{KJ}}$
Then the loss in kinetic energy during the collision is obtained as $\Delta K = {K_a} - {K_b}$
On substituting the values obtained for ${K_b}$ and ${K_a}$ in the above relation we get, $\Delta K = \dfrac{{20}}{3} - 20 = 13.333{\text{KJ}}$ or $\Delta K = 1.33 \times {10^4}{\text{J}}$
$\therefore $ the loss in kinetic energy during the collision is $\Delta K = 1.33 \times {10^4}{\text{J}}$ .

Hence the correct option is D.

Note: Here, the collision is perfectly inelastic as the cars stick together after the collision. The kinetic energy is not conserved for such collisions. While substituting values of physical quantities in an equation make sure that the quantities are expressed in their respective S.I. units. If not necessary conversions of the unit must be done. Here, the masses of the cars are expressed in metric tonnes. During substitution, this is converted to kilograms using the unit conversion relation given by $1{\text{ metric tonne}} = 1000{\text{kg}}$ .