Motion In Two Dimensions Question 4
Question 4 - 29 January - Shift 2
A particle of mass $100 g$ is projected at time $t=0$ with a speed $20 ms^{-1}$ at an angle $45^{\circ}$ 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
nan
(Take $g=10 ms^{-2}$ )
Show Answer
Answer: (800)
Solution:
Formula: Angular momentum
Use $\Delta L=\int_0^{t} \tau dt$
$L_0=\int_0^{2} mg(v_x t) dt$
$=mg_x \frac{t^{2}}{2}=(0.1)(10)(10 \sqrt{2}) \frac{2^{2}}{2}$
$=20 \sqrt{2}$
$=\sqrt{800} kg m^{2} / s$
