A particle of mass m moves uniformly with a speed v along a circle of radius r- A and B are two points on the circle- such that the arc AB subtends an angle theta the center of the circle- The magnitude of change in momentum as the particle moves from A to B is given by-2 sin -dfrac-theta-2-2 cos -dfrac-theta-2-zero2 tan -dfrac-theta-2-

Question

Toppr Admin · Accepted Answer

Assume particle is at lowest point of the circle

$P_{i} = m v^i$

After it rotates $θ$ angle

$P_{f} = m v cos θ^i + m v sin θ^j$

Change in momentum $= P_{f} - P_{1} =$ $m v cos θ^i + m v sin θ^j - m v^i$

$= m v (- 1 + cos θ)^i + m v sin θ^j$

$| Δ P | = m v \sqrt{(- 1 + cos θ)^{2} + {sin}^{2} θ}$

$| Δ P | = m v \sqrt{1 + {cos}^{2} θ - 2 cos θ + {sin}^{2} θ}$

$| Δ P | = 2 m v sin (\frac{θ}{2})$

Assume particle is at lowest point of the circlePi=mv^iAfter it rotates θ angle Pf=mvcosθ^i+mvsinθ^jChange in momentum =Pf−P1= mvcosθ^i+mvsinθ^j−mv^i =mv(−1+cosθ)^i+mvsinθ^j|ΔP|=mv√(−1+cosθ)2+sin2θ |ΔP|=mv√1+cos2θ−2cosθ+sin2θ|ΔP|=2mvsin(θ2)