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प्रश्न
The below figure shows the position-time graph of a particle of mass 4 kg.
- What is the force on the particle for t < 0, t > 4 s, 0 < t < 4 s?
- What is the impulse at t = 0 and t = 4 s? (Consider one-dimensional motion only.)
उत्तर
- For t < 0
For t<0, the graph of position versus time is denoted as BO, which indicates that the displacement of the particle is zero, i.e., the particle is stationary at the origin. Consequently, the force exerted on the particle must be zero. - For t > 4 s
For t>4 s, the position-time graph segment AC runs parallel to the time axis, indicating that the particle maintains a position 3 metres from the origin, implying it is stationary. Therefore, the force acting on the particle is zero. - For 0 < t < 4
Between 0 < t < 4 s, the position-time graph labelled OA displays a constant slope, which indicates that the velocity of the particle remains constant during this interval, meaning the particle has zero acceleration. Consequently, the force on the particle must be zero.
- For t < 0
- At t = 0
Impulse = Change in momentum
= mv – mu
Mass of the particle, m = 4 kg
Initial velocity of the particle, u = 0
Final velocity of the particle, `v = 3/4 "m/s"`
∴Impulse = `4(3/4 - 0) = 3 kg "m/s"`
At t = 4 s
Initial velocity of the particle, `u = 3/4 "m/s"`
Final velocity of the particle, v = 0
∴ Impulse = `4(0-3/4)` = -3 kg m/s
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