a block of mass moving on a smooth horizontal plane with a velocity v collides with a stationary block of mass m at the back of which spring of spring constant k is attached as shown in the figure
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HEY MATE
HERE IS YOUR ANSWER
•FOR BLOCK MOVING WITH VELOCITY V
KINETIC ENERGY INITIAL=1/2MV²
•FOR SPRING
WORK DONE FINAL=1/2KX²
K. Ei+W.Df-W.Di=K. Ef
AS W.Di =0
K. Ef=0
k.ei=w. d
1/2MV²=1/2KX²
MV²=KX²
X=V√M/K
SO THE EXTENSION IN WIRE IS
V√(M/K)
HERE IS YOUR ANSWER
•FOR BLOCK MOVING WITH VELOCITY V
KINETIC ENERGY INITIAL=1/2MV²
•FOR SPRING
WORK DONE FINAL=1/2KX²
K. Ei+W.Df-W.Di=K. Ef
AS W.Di =0
K. Ef=0
k.ei=w. d
1/2MV²=1/2KX²
MV²=KX²
X=V√M/K
SO THE EXTENSION IN WIRE IS
V√(M/K)
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