Strike it hard

A hammer of mass M M falls from a height l l to strike a pile of mass m m into ground. Pile penetrates a distance d d in a single blow. Find the resistance to penetration given by ground in N e w t o n s Newtons

Details and Assumptions

  • Assume collision to be inelastic.
  • Assume both the masses to be point masses.
  • M = 0.5 kg , m = 5 kg , l = 22 cm , d = 5 cm , g = 10 m / s 2 . M=0.5 \text{kg} , \space m=5\text{kg} , \space l=22\text{cm}, \space d = 5\text{cm} , \space g=\SI[per-mode=symbol]{10}{\meter\per\second\squared}.


The answer is 57.

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1 solution

Kushal Patankar
Mar 5, 2015

Velocity of hammer at the time of contact will be 2 g l \sqrt{2gl} Apply law of conservation of momentum ( M + m ) v = M 2 g l (M+m)v = M \sqrt{2gl} v = M 2 g l ( M + m ) \Rightarrow v= \frac{ M \sqrt{2gl}}{(M+m)}

We know that work done by a force equals force times displacement.

And of course everyone knows about work energy theorem

R × d ( M + m ) g d = 1 2 ( M + m ) ( M 2 g l ( M + m ) ) 2 R \times d - (M+m)gd = \frac{1}{2} (M+m) \Big(\frac{ M \sqrt{2gl}}{(M+m)}\Big) ^ {2} R = M 2 g l ( M + m ) d + ( M + m ) g \Rightarrow R = \frac{M^{2} g l}{(M+m)d} + (M+m)g Plug in the values now.

Excellent Problem And Solution!!!!!

Prakhar Bindal - 6 years, 3 months ago

Nice problem and amazing solution! Kindly mention that collision was "perfectly" inelastic. I got stuck up there.

Aditya Kumar - 5 years, 6 months ago

i insist, you must write in the problem that resistance is constant otherwise you can't even think of usig the word R in the solution ! @Kushal Patankar

A Former Brilliant Member - 4 years, 6 months ago

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