The following questions appeared in Kerala Engineering Entrance 2007 question paper:

**A shell at rest at the origin explodes into three fragments of masses 1 kg, 2 kg and ‘m’ kg. The 1 kg and 2 kg pieces fly off with speeds of 5 ms ^{–1} along the X- axis and 6**

**ms**

^{–1}along the Y-axis respectively. If the m kg piece flies off with speed 6.5**ms**

^{–1}, the total mass of the shell must be**(a) 4 kg (b) 5 kg (c) 3.5 kg (d) 4.5 kg (e) 5.5 kg **

The initial momentum of the shell is zero. Explosion of the shell involves internal forces only. Since there are no external forces, the total momentum of all the fragments of the shell taken together must be zero.

The momentum of the 1 kg fragment is 5 kg ms^{–1} and that of the 2 kg fragment is 12 kg ms^{–1}. Since these are at right angles, the net momentum of these two fragments taken together is √(5^{2} + 12^{2}) = 13 kg ms^{–1}. The momentum of the m kg fragment must be equal and opposite to this (to make the net momentum of the three fragments zero).

Therefore, m×6.5 = 13, from which m = 2 kg.

The mass of the shell is therefore 1 + 2 + 2 = 5 kg.

The following question is simple, but many among you will have certain doubts about it:

**A uniform rope of length ‘L’ is resting on a smooth horizontal surface. If it is pulled at one end by applying a horizontal force ‘F’ parallel to its length, what will be the tension in the rope at a distance ‘d’ from the other end?**

**(a) F (b) F/d (c) FL/d (d) Fd (e) Fd/L**

When you work out questions on bodies connected by strings, you will assume that the string is light and therefore the tension in a segment of the string connecting two bodies will be the same everywhere. The acceleration of the system in such cases will be determined by the masses of the connected bodies only.

In the above problem, there are no bodies connected to the rope and hence the acceleration ‘a’ of the rope is determined by the mass of the rope only.

Therefore, a = F/ mL where ‘m’ is the mass per unit length of the rope.

The tension ‘T’ in the rope at the distance ‘d’ from the other end is responsible for pulling the portion of length ‘d’ of the rope. Therefore we have

T = mass × acceleration = md × F/mL = Fd/L

The following question on connected bodies is a popular one and you are expected to know how to solve it:

**Three blocks of masses m _{1}, m_{2} and m_{3} are connected by light inextensible strings and carried by a light frictionless pulley as shown. If (m_{1}+m_{2}) > m_{3}, the tension in the string connecting m_{1} and m_{2} is **

**(a) 2m _{1}m_{3}g/ (m_{1}+m_{2}+m_{3}) **

**(b) 2(m _{1}+m_{2})m_{3}g/(m_{1}+m_{2}+m_{3})**

**(c) m _{1}g **

**(d) (m _{1}+m_{2}) **

**(e) 2m _{1}m_{2}m_{3}g/ (m_{1}+m_{2}+m_{3}) **

Note that the tension in the string connecting m_{1} and m_{2} is different from the tension in the string connecting m_{2} and m_{3}.

The common acceleration ‘a’ of the system of masses is the ratio of the net driving force to the total mass moved.

Therefore, a = (m_{1}+m_{2}** **– m_{3})g/(m_{1}+m_{2}+m_{3})

The tension (T) in the string connecting m_{1} and m_{2} is produced by the apparent weight ( as in the case of motion in a lift) of the mass m_{1} which moves downwards with acceleration ‘a’.

Therefore, T = m_{1}(g –** **a)** = **m_{1}[g – (m_{1}+m_{2}** **– m_{3})g/(m_{1}+m_{2}+m_{3})]

= m_{1}g[1– (m_{1}+m_{2}** **– m_{3})/(m_{1}+m_{2}+m_{3})]

= 2m_{1}m_{3}g/ (m_{1}+m_{2}+m_{3})

You will find more questions (with solution) in this section by clicking on the label ‘Newton’s laws’ below this post or on the side of this page.

Similar multiple choice questions with solution can be found at physicsplus: Multiple Choice Questions on Newton’s Laws of Motion

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