three blocks x,y and z of masses m, 2m and 3m respectively are placed on horizontal smooth plane.Block x with speed 18m/s collides and sticks with block y.Both objects collides and sticks with block z.Together its move with common velocity,v.
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Conservation of linear momentum.
Mi*Vi = Mf*Vf
(1)*(18m/s) + (2)*(0m/s) + (3)*(0m/s) = (1+2+3)*(?)
(18m/s)/(6)=(?)=3m/s
Final velocity = 3m/s
Mi*Vi = Mf*Vf
(1)*(18m/s) + (2)*(0m/s) + (3)*(0m/s) = (1+2+3)*(?)
(18m/s)/(6)=(?)=3m/s
Final velocity = 3m/s
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Assuming a frictionless surface and total energy is conserved; then the Kinetic Energy (KE) of block X= KE of the system of blocks X+Y+Z.
Since KE is proportional to mv², we can solve for the after collision velocity (v) of the X+Y+Z system of mass M.
ΣM=m + 2m + 3m = 6m,
m (18)²= 6m(v)² solving for v
v²= (18)²/6
v= 7.35 m/s
Since KE is proportional to mv², we can solve for the after collision velocity (v) of the X+Y+Z system of mass M.
ΣM=m + 2m + 3m = 6m,
m (18)²= 6m(v)² solving for v
v²= (18)²/6
v= 7.35 m/s