Problem (IIT JEE 2002): Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the centre of mass is,
Problem (IIT JEE 1994):
The magnitude of the force (in newton) acting on a body varies with time $t$ (in microseconds) as shown in the figure. AB, BC and CD are straight line segments. The magnitude of the total impulse of the force on the body from $t={4}\;\mathrm{\mu s} $ to $t={16}\;\mathrm{\mu s} $ is _______ N s.
Solution: The impulse, $I=\int F\mathrm{d}t$, is the area under the $F\text{-}t$ curve from $t={4}\;\mathrm{\mu s}$ to $16\;\mathrm{\mu s}$. This area is given by, \begin{align} I=(2\times200+\tfrac{1}{2}\times 2\times 600+\tfrac{1}{2}\times 10\times 800){10}^{-6}=5\times{10}^{-3}\;\mathrm{N\,s}.\nonumber \end{align}