Practice portal › Work, Energy and Power › Springs and Elastic Potential Energy
Asked in JEE Main 11th Jan 1st Shift 2019 · Springs with friction or impact
Speed just before impact: v=√2gh=√2000 m/s, so v²=2000.
The body sticks (perfectly inelastic): 1× v=4v'⇒ v'²=(v²)/(16)=125.
Kinetic energy after sticking: 1/2×4×125=250 J.
Compression: 1/2kx²≈250 J (the extra 4g x≈0.8 J of gravity work over 2 cm is negligible): x²=(500)/(1.25×10⁶)=4×10⁻⁴.
x=0.02 m =2 cm.
Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.
Practise this with a timer