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A mass m which is attached to a spring constant k, oscillates on a horizontal table with amplitutde A.At the instant when the spring is stretched by 3^1/2A/2(Aroot3/2),a second mass m is dropped vertically onto the original mass and immediately sticks to it.What is the amplitude of the resulting motion

2 years ago

Amresh
13 Points

As there is no net external force acting on the system, the linear momentum will remain conserved.
So, mv=2mv'
v'=v/2
Now, Energy will be conserved
1/2k2=1/2*2*m*(v')2+1/2*k*3A2/4
As v=
8 months ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions