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Second Law of Motion

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Question

The figure represents two bodies of masses 10 kg and 20 kg and moving with an initial velocity of $10 m s^{- 1}$ and $5 m s^{- 1}$ , respectively. They are colliding with each other, After collision, they are moving with velocities $12 m s^{- 1}, 4 m s^{- 1}$ , respectively. The time of collision is 2s. Then calculate $F_{1}$ and $F_{2}$ .

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Solution

Consider object of masses 10kg and 20 kg as A and B object A,
Mass of object A,
$m_{1} = 10 k g$
Initial velocity of object A, $u_{1} = 10 m s^{- 1}$
Final velocity of A, $v_{1} = 12 m s^{- 1}$
Mass of object B, $m_{2} = 20 k g$
Initial velocity of object B, $u_{2} = 5 m s^{- 1}$
Final velocity of B, $v_{2} = 4 m s^{- 1}$ Formula $F = \frac{m (v - u)}{t}$
For object A, $F_{1} = \frac{m_{1} (v_{1} - u_{1})}{t}$
$F_{1} = \frac{10 (12 - 10)}{2} = 10 N$
For object B, $F_{2} = \frac{m_{2} (v_{2} - u_{1})}{t}$
$F_{2} = \frac{20 (4 - 5)}{2} = - 10 N$
As $F_{1} = 10 N$ and $F_{2} = - 10 N$ which proves Newton's third law of motion (i.e,) conservation of momentum

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