A particle of mass M originally at rest is subjected to a force whose direction is constant but magnitude varies with time according to the relation
F = F₀[ 1 - ( (t - T)/T )² ]
Where F₀ and T are constants. The force acts only for the time interval 2T. The velocity v of the particle after time 2T is :

Solution

Physics sir · 4 steps

The force keeps changing, so integrate it over time: that impulse is the momentum the particle gains.

1. Impulse gives momentum. Newton's second law says F = dp/dt, so the change in momentum is ∫ F dt. The particle starts from rest, so Mv = ∫₀^(2T) F dt.

   Mv = F₀∫₀^(2T)[1-((t-T)/T)²]dt

2. Substitute. Let u = (t-T)/T, so dt = T du. When t = 0, u = -1, and when t = 2T, u = 1. The integral becomes F₀ T∫₋₁¹(1-u²) du.
3. Evaluate. ∫₋₁¹(1-u²) du = [u - u³/3]₋₁¹ = 2/3 - (-2/3) = 4/3. So the impulse is 4/3F₀ T.
4. So. Mv = 4F₀T/3, which gives v = 4F₀T/3M, option (c).

The idea to keep. When the force changes with time, do not use F = Ma with one fixed value. Impulse ∫ F dt equals the change in momentum, and it is the area under the force-time graph.

Answer from the NTA final answer key. Solution by PrepWiser.