1And for 3rd
by work k.E. thm,
1/2mv2-0=WT+Wgravity
So, Wtension=mgd+1/2mv2
1. a ball is allowed to fall from a height of 10 metre. if there is 40% loss of energy due to impact then after one impact ball will go upto
a, 10 b. 8
c.6 d. 4
2. a particle is projected with kinetic energy k. its range is R. it will have minimum kinetic energy after travelling horizontal distance
a. R/4 b. R/2
c. 3R/4 d. R
3. in the shown figure if block moves upward with constant velocity v, then work done by tension in rope in displacing it upward by distance d will be
a. mgd+1/2mv2 b. mgd
c. 1/2mv2
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UP 0 DOWN 0 0 10
10 Answers
1
106Q1.. v = √2gh so initial KE = m*2gh/2 = 10mg
after collision . mv2/2 = 0.6*10mg = 6mg
So, v2 = 12g
==> v= √12g
max height = v2/2g = 12g/2g = 6 m
Q2. KE = m(Vx2 + Vy2)/2
KE is minimum when Vy =0 as Vx remains constant and Vy = 0 when particle is at max. height.. i.e. horizontal distance = R/2