**6**
AKHIL
**·**2011-04-05 05:07:18
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
**·**2011-04-05 07:23:09
Why don't you solve it Akhil ? :-p

**1**
varun.tinkle
**·**2011-04-05 09:11:50
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
**·**2011-04-05 10:34:57
Its not a perfectly elastic collision. Final total energy will contain energy produced due to vibration, sound etc...

**1**
swordfish
**·**2011-04-05 10:43:19
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=âˆš2gh

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.

**30**
Ashish Kothari
**·**2011-04-06 00:38:15
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]

**30**
Ashish Kothari
**·**2011-04-06 00:42:32
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
**·**2011-04-06 01:06:50
@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!

**6**
AKHIL
**·**2011-04-07 00:04:31
i believe more in giving hints than solving each and evry prob!!!

:P

**1**
swordfish
**·**2011-04-07 01:04:28
**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
**·**2011-04-07 21:09:56
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
**·**2011-04-08 01:16:39
Energy loss is certain. There is no need to specify it. You never see an elastic collision between a hammer and a nail.

**Anonymous**
**·**2018-09-29 16:33:29
While applying work energy theorem wouldn't v consider (m+M )gÃ—d {work done by gravity}??