Question
1. A mass "m" moving with a velocity "u" hits a surface at an angle "Q" with the normal at the point of hitting . How much force does it exerts, if no energy is lost ?
2. A thin cicular loop of radius "R" rotates about its vertical diameter with angular frequency "W" . show that a small bead on the wire loop remains at its lowermost point for W ≤ (root of g/R) . what is the angle made by the radius vector joining the center to the bead with the vertical downwards direction for W= ( roots of 2g/R) . neglect friction .
3. A rear side of a truck is open and a box of 40 kg is placed 5m away from the open end . the coefficient of friction b/w box and surface in 0.15 on a straight road , the truck starts from rest and acclerates with 2m/s2. at what distance from the starting point does the box fall off the truck .
4. Straight from rest , a mass "m" slides down on inclined plane "Q" in a time "n times " the time to slide down the same lenght in absence of friction . Find the coefficient of friction .
1) angle is measured from vertical so vertical component of velocity = ucosθ
and since no energy is lost so returning velocity in vertical direction will be same
so impulse = (mucosθ) - (-mucosθ)
force cant be calculated without time of impact.
2)
at the position given in diagram mrω2cosθ = mgsinθ for equilibrium
so rω2cosθ = gsinθ and r = Rsinθ
so Rsinθω2cosθ = gsinθ so Rω2cosθ = g or ω = √[g/Rcosθ] so ω ≥ √[g/R]
thus for W ≤ (root of g/R), bead will remain at the lowermost point.
for second part compare W= ( roots of 2g/R) and ω = √[g/Rcosθ]
3) pseudo in opposite direction = ma
frictional force in the direction of motion of truck = 0.15mg
so equation of motion of box in opposite direction is ma - 0.15mg = ma'
so 2-1.5 = a' or a' = 0.5
distance = 5m
so time taken upto fall of box can be obtained by s = 1/2 a' t2
after calculating t, use again s = 1/2 (2) t2 for finding distance travelled by truck.
4) in first case with friction s = 1/2 (gsinθ-μgcosθ) (nt)2
in second case s = 1/2 (gsinθ) t2
now solve.