in the quiz the question was:Two blocks of the masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block. The velocity of centre of mass is 30 m/s 20 m/s 10 m/s 5 m/s Incorrect Correct Answer: C According to principle of conservation of motion MV=(m+M)Vcm. Therefore, v=m1v1+m2v2/m1+m2=m(2v)+m(-v)/m+m=v/2(10*14/10+4)=10 m/s Mgh/x Vcm=MV/m+m=10*4/10+4=10 m/s.why were the values-(2v) & (-v)put in the step: Vcm=m(2v)+m(-v)/m+m.please explain in an elaborated manner.
Asked by Ayushi Tiwari | 9th Nov, 2013, 12:37: PM
There is no need to include negative values anywhere.
The solution is solved simply as below:
The impulse velocity is given to heavier mass m1 and not to m2, so v1 = 14 m/s and v2 = 0.
Using conservation of momentum, we have
Answered by Romal Bhansali | 11th Nov, 2013, 11:36: AM
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