First slide

Second law

Question

A body of 2 kg has an initial speed 5 ${ms}^{- 1}$ . A force acts on it for some time in the direction of motion. The force time graph is shown in figure. The final speed of the body is

Moderate

A

9.25 ${ms}^{- 1}$
B

5 ${ms}^{- 1}$
C

14.3 ${ms}^{- 1}$
D

4.25 ${ms}^{- 1}$

Solution

Impulse = Change in momentum = $m (v_{2} - v_{1}) - - - - (i)$
Again impulse = Area between the graph and time axis
= $\frac{1}{2} \times 2 \times 4 + 2 \times 4 + \frac{1}{2} (4 + 2.5) \times 0.5 + 2 \times 2.5$
= 4+8+1.625+5 = 18.625----------(ii)
From (i) and (ii), m( $v_{2} - v_{1}$ ) = 18.625
$\Rightarrow v_{2} = \frac{18.625}{m} + v_{1} = \frac{18.625}{2} + 5 = 14.3 m / s$

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