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
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?
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
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.....
Energy loss is certain. There is no need to specify it. You never see an elastic collision between a hammer and a nail.
While applying work energy theorem wouldn't v consider (m+M )g×d {work done by gravity}??
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...
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)
Its not a perfectly elastic collision. Final total energy will contain energy produced due to vibration, sound etc...
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=M^{2}/(M+m)d
Now the total opposing force on the system = F + (M+m)g = M^{2}gh/(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.
Varun , my approach was similar, but what Swordfish has done is right. The right answer is indeed M^{2}gh(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]
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]
@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!