Hammer problem!

Q. A hammer of mass M falls from a height h repeatedly to drive a pile of mass m into the ground. The hammer makes the pile penetrate in the ground to a distance d in single blow. Opposition to penetration is given by?

14 Answers

22
Ashish Kothari ·

Thanks! [1]

6
AKHIL ·

i believe more in giving hints than solving each and evry prob!!!
:P

1
swordfish ·

i believe more in giving hints than solving each and evry prob!!!

I think it should be like - 'I believe more in giving hints to be on the safe side rather than solving each and evry prob wrongly!!!' [4]

No Offence

1
varun.tinkle ·

Well nice answer
But on the question its very unclear actually confisuing ... I am dammn sure such confusing questions wont come.... in any exam . All the physics we study are based on the assumptions of idealism.... so it should have been clearly specifed that apart from the opp. force there can be enrgy losss. I beleive so.....

1
swordfish ·

Energy loss is certain. There is no need to specify it. You never see an elastic collision between a hammer and a nail.

Anonymous ·

While applying work energy theorem wouldn't v consider (m+M )g×d {work done by gravity}??

6
AKHIL ·

i think it means the force imparted by the pile on to the hammer , rite??

so , its simple , find the impulse.........and then force can be found out...

1
swordfish ·

Why don't you solve it Akhil ? :-p

1
varun.tinkle ·

Well i guess.... work power energy gives a nice and convininet explaination....

now first...
initially taking 0 P.E. to be at the ground level....

initially total energy=Mgh
and finally total enrgy= -(M+m)gd
thereofre change=F.s where s= d

Assumption... The masses are point masses

(Ashish could u pls post problems by the proffesor who coaches the indian ipho team for our practises)

1
swordfish ·

Its not a perfectly elastic collision. Final total energy will contain energy produced due to vibration, sound etc...

1
swordfish ·

I am posting the solution which I rarely do :D
Just before and after the collision, momentum of the two masses is conserved (approximately).
Mv=(M+m)v'
v'=Mv/(M+m), where v=√2gh

Now applying work-energy theorem,

0.5(M+m)(0 - v'2)=-F*d, where F is the resistive force

F=M2/(M+m)d

Now the total opposing force on the system = F + (M+m)g = M2gh/(M+m)d + (M+m)g

Here I made an assumption that the hammer doesnot rebound before travelling a distance 'd' and comes to rest together with the pile.

22
Ashish Kothari ·

Varun , my approach was similar, but what Swordfish has done is right. The right answer is indeed M2gh(M+m)d + (M+m)g.

@Swordfish - One question, why is the term (M+m)g included in the total opposing force on the system? [7]

22
Ashish Kothari ·

And by the way, are you confusing me with Asish bhaiya (like most people out here do), who is currently in IIT-K? I don't know the professor who you are talking about! [3]

71
Vivek @ Born this Way ·

@ashish

(M+m)g will also also because that's the gravitational force of attraction on the combined bodies. The ground has to oppose both the impulse and weight. That's all!

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