Question

A truck of mass \[10\] metric ton runs at \[3{\rm{ m}}{{\rm{s}}^{ - 1}}\] along a level track and collides with a loaded truck of mass \[20\] metric ton, standing at rest. If the trucks coupled together, the common speed after collision is
A. \[1{\rm{ m}}{{\rm{s}}^{ - 1}}\]
B. \[10{\rm{ m}}{{\rm{s}}^{ - 1}}\]
C. \[0.5{\rm{ m}}{{\rm{s}}^{ - 1}}\]
D. \[0.3{\rm{ m}}{{\rm{s}}^{ - 1}}\]

Answer 1

Hint: To solve this problem we will be using the concept of conservation of momentum before collision and after collision which states that the momentum is always conserved.

 Complete step by step answer:
Assume,
\[{m_1}\] is the mass of truck travelling with speed a of \[3{\rm{ m}}{{\rm{s}}^{ - 1}}\].
\[{m_2}\] is the mass of the truck which is at rest.

The concept of conservation of momentum states that the summation of momentum of two colliding bodies before collision is equal to the summation of momentum of these two bodies after collision.

The truck of mass \[10\] metric ton running at a speed of \[3{\rm{ m}}{{\rm{s}}^{ - 1}}\] along a level track is colliding with the truck of mass \[20\] metric ton which is at the rest position.

Using the concept of conservation of momentum we can state that the summation momentum of both trucks before collision is equal to the summation of their momentum after collision.

It is given that the truck of mass \[20\] metric ton is at rest before collision so its momentum before collision will be zero.

Writing the equation of conservation of momentum for both trucks.
\[{m_1}{u_1} + {m_2}{u_2} = mv\]……(1)

Here \[{u_1}\] and \[{u_2}\]are velocities before collision of trucks with mass \[{m_1}\] and \[{m_2}\] respectively. Also, u and m are velocity and mass of the coupled unit after collision.

After collision the mass of the coupled unit will be the summation mass \[{m_1}\] and \[{m_2}\].

Substitute \[10{\rm{ to}}{{\rm{n}}^3}\] for \[{m_1}\], \[10{\rm{ to}}{{\rm{n}}^3}\] for \[{m_2}\], \[3{\rm{ m}}{{\rm{s}}^{ - 1}}\] for \[{u_1}\], 0 for \[{u_2}\] and \[30{\rm{ to}}{{\rm{n}}^3}\] for m in equation (1).
\[\begin{array}{l}
\left( {10{\rm{ to}}{{\rm{n}}^3}} \right)\left( {3{\rm{ m}}{{\rm{s}}^{ - 1}}} \right) + \left( {20{\rm{ to}}{{\rm{n}}^3}} \right) \cdot 0 = \left( {30{\rm{ to}}{{\rm{n}}^3}} \right)v\\
v = 1{\rm{ m}}{{\rm{s}}^{ - 1}}
\end{array}\]

Therefore, the velocity of the coupled unit of both trucks after collision is \[1{\rm{ m}}{{\rm{s}}^{ - 1}}\]

So, the correct answer is “Option A”.

Note:
While writing the equation of conservation of momentum, do not treat the trucks as different bodies as it is given that they are coupled after collision.