A particle of mass 100 g is projected at time t=0 with a speed 20 ms -1 at an angle 45° to the horizontal as given in the figure. The magnitude of the angular momentum of the particle about the starting point at time t=2 s is found to be √ K kg m 2 / s. The value of K is (Take g=10 ms -2 )

Question

A particle of mass 100 g is projected at time t=0 with a speed 20 ms -1 at an angle 45° to the horizontal as given in the figure. The magnitude of the angular momentum of the particle about the starting point at time t=2 s is found to be √ K kg m 2 / s. The value of K is (Take g=10 ms -2 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Use Δ L =∫ limits0t τ d t L 0=∫ limits02 mg ( v x t ) dt = mgv x ( t 2/2)=(0.1)(10)(10 √2) (22/2) =20 √2 =√800 kg m 2 / s

Q. A particle of mass 100g is projected at time t=0 with a speed 20ms−1 at an angle 45∘ to the horizontal as given in the figure. The magnitude of the angular momentum of the particle about the starting point at time t=2s is found to be K​kgm2/s. The value of K is __ (Take g=10ms−2 )

Use ΔL=0∫t​τdt L0​=0∫2​mg(vx​t)dt =mgvx​2t2​=(0.1)(10)(102​)222​ =202​ =800​kgm2/s

Use $Δ L = 0 \int t τ d t$
$L_{0} = 0 \int 2 m g (v_{x} t) d t$
$= m g v_{x} \frac{t ^{2}}{2} = (0.1) (10) (102) \frac{2 ^{2}}{2}$
$= 202$
$= 800 k g m^{2} / s$