First slide

Second law

Question

In the figure given below, the position - time graph of a particle of mass 0.1 kg is shown. The impulse at t = 2 sec is

Moderate

A

0.2 $kg m \sec^{- 1}$
B

-0.2 $kg m \sec^{- 1}$
C

0.1 $kg m \sec^{- 1}$
D

-0.4 $kg m \sec^{- 1}$

Solution

Velocity between t = 0 and t = 2 sec
$\Rightarrow v_{1} = \frac{dx}{dt} = \frac{4}{2} = 2 m / s$
Velocity at t = 2 sec, $v_{f} = 0$
Impulse = Change in momentum = m ( $v_{f} - v_{f}$ )
0.1(0-2) = -0.2 $kg m \sec^{- 1}$

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