First slide

Law of conservation of momentum and it's applications

Question

The adjacent figure shows the position graph of one dimensional motion of a particle of mass 4kg. The impulse at t=0 and t=4s is given respectively as:

Moderate

A

$0, 0$
B

$0, - 3 kg {ms}^{- 1}$
C

$+ 3 kg {ms}^{- 1}, 0$
D

$+ 3 kg {ms}^{- 1}, - 3 kg \cdot {ms}^{- 1}$

Solution

$\begin{array}{l} t < 0, v_{i} = 0 and t > 0, v_{f} = \frac{3}{4} m / s \\ ∴ Impulse = m (v_{f} - v_{i}) = 4 (\frac{3}{4} - 0) = 3 kg {ms}^{- 1} \\ Little before 4 second v = \frac{3}{4} m / s, little after 4 second, velocity becomes zero . \\ Therefore, Impulse = m (v_{f} - v_{i}) = 4 (0 - \frac{3}{4}) = - 3 kg {ms}^{- 1} \end{array}$

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