Question

A truck is moving on a frictionless surface with a uniform velocity \[{\rm{10m}}{{\rm{s}}^{{\rm{ - 1}}}}\]. A leak occurs in the water tank of the truck at the rate of \[{\rm{2kg}}{{\rm{s}}^{{\rm{ - 1}}}}\]. What is the speed of the truck after \[{\rm{50s}}\], if the mass of the truck is \[{\rm{100kg}}\] and the mass of water in the truck initially was \[{\rm{100kg}}\]?
A. \[{\rm{20m}}{{\rm{s}}^{{\rm{ - 1}}}}\]
B. \[{\rm{10m}}{{\rm{s}}^{{\rm{ - 1}}}}\]
C. \[{\rm{15m}}{{\rm{s}}^{{\rm{ - 1}}}}\]
D. None of these

Answer 1

Hint: Before we proceed with the problem, it is important to know about the law of conservation of momentum and uniform velocity. The law of conservation of momentum states that before and after the collision the total momentum of a system should be conserved. Now we can solve the problem step by step as follows.

Formula Used:
By the law of conservation of momentum
Initial momentum= final momentum
i.e., \[{m_1}{u_1} = {m_2}{u_2}\]
Where,
\[{m_1}\] is the initial mass.
\[{u_1}\] is the initial velocity.
\[{m_2}\] is the final mass.
\[{u_2}\] is the final velocity.

Complete step by step solution:
Mass of the truck \[{\rm{ = 100\,kg}}\]
Total mass of the system= mass of the truck +mass of water in the truck
Total mass of the system \[ = (100 + 100){\rm{ = 200kg = }}{{\rm{m}}_1}\]
The water leaks at the rate of \[{\rm{2kg}}{{\rm{s}}^{{\rm{ - 1}}}}\].
So, after 50 sec
\[m = 50 \times 2 = 100\,kg\]
Now, the final mass of the truck \[ = 200 - 100 = 100\,kg\]

From the formula we will find the value of \[{u_2}\].
\[{m_1}{u_1} = {m_2}{u_2}\]
\[\Rightarrow {u_2} = \dfrac{{{m_1} \times {u_1}}}{{{m_2}}}\]
\[\Rightarrow {u_2} = \dfrac{{200 \times 10}}{{100}}\]
\[\therefore {u_2} = 20\,m{s^{ - 1}}\]
Therefore, the speed of the truck is \[{\rm{20m}}{{\rm{s}}^{{\rm{ - 1}}}}\].

Hence, option A is the correct answer.

Note: Let’s see an example for the law of conservation of linear momentum. When a bullet is shot from a gun, both the bullet and the gun are at first very still i.e., the total momentum before firing is zero. Then the bullet gains forward momentum when it gets discharged. As per the law of conservation of momentum, the gun gets regressive momentum. Due to this, the bullet of mass m is terminated with velocity v. The gun of mass M gains a backward velocity u. Then, the total momentum becomes zero. Thus, the total momentum after firing is also zero.