Please solve this
Q.21. A loaded 20,000 kg coal wagon is moving on a linear track with a 8 m/s. Suddenly 8000 kg of coal is dropped out of wagon. The final speed of the wagon is
(1) 6 m
(2) 8 m
(3) m/s
(4) m/s
M1 = 20,000; v1 = 6 m/s ; m2=20000-5000= 15000 ; for dropped coal m = 5000 v = 6 m/s ( it is dropped with same speed)
Now by conservation of linear momentum,
M1v1 = m2v2 - mv
20000(6) = 15000(v2) - 5000(6)
V = 6 m/s