Law of conservation of momentum and it's applications

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Question

A box of mass m = 10 kg resting on a frictionless horizontal ground is acted upon by a horizontal force F, which varies as shown. Find speed of the particle (in m/s) when the force ceases to act.

Moderate

Solution

The area under force-time graph gives the impulse. Hence impulse acting on the box.
$\vec{J} = \int_{t_{i}}^{t_{f}} \vec{F} dt = Area under the graph = \frac{1}{2} \times 100 \times 2 = 100 N - s$
Using impulse-momentum equation:
$m {\vec{v}}_{f} = m {\vec{v}}_{i} + \int_{t_{i}}^{t_{f}} \vec{F} dt or m {\vec{v}}_{f} = m {\vec{v}}_{i} + \vec{J}$
$10 \times v = 10 \times 0 + 100 \Rightarrow v = 10 m / s$

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