A particle of mass 20 g is released with an initial velocity 5 m/s along a frictionless curve from the point A. The point A is at a height h=10 m above the lowest point B of the curve, and the reference point O lies on the vertical through B, a further a=10 m above the level of A. When the particle…

Question

A particle of mass 20 g is released with an initial velocity 5 m/s along a frictionless curve from the point A. The point A is at a height h=10 m above the lowest point B of the curve, and the reference point O lies on the vertical through B, a further a=10 m above the level of A. When the particle reaches the point B, its angular momentum about O will be (take g=10 m/s²)

Accepted Answer

(1) 6 kg m²/s. First find how fast the particle is moving at B. The curve is frictionless, so energy is conserved over the drop h=10 m. v_B²=v_A²+2gh=25+2(10)(10)=225, so v_B=15 m/s. At B the particle is at the bottom of the curve, so its velocity is horizontal. O lies directly above B, a distance a+h=10+10=20 m up. The perpendicular distance from O to the horizontal line of motion at B is therefore the full 20 m. L=mv_Bd=(0.02)(15)(20). L=6 kg m²/s.

Step-by-step solution

Why the other options are wrong