Mechanics - falling ball and impulse

[cupid.callin]

Homework Statement

Hi all

The Attempt at a Solution

First of all,
can someone tell me how to get the eqn in hint?

And when i try solving it using the hint ...

of ball

(velocity of separation) = e(velocity of approach)
v' = e √(2gh) = √(gh/2)

let the time of collision is t

Impulse, I_Ball = Δp = m(v' - v) = m ( -√(gh/2) )

I_Block = μ I_Ball = -0.2 m√(gh/2)

I_Block = m Δv
Δv = 0.1 √(2gh)

But answer is (D)

 

[tiny-tim]

hi cupid.callin!

your I_ball is wrong

 

[ashishsinghal]

Impulse, I_Ball = Δp = m(v' - v) = m ( -√(gh/2) )

But answer is (D)

Impulse, I_Ball = Δp = m(v' + v)
the answer D is correct!

 

[cupid.callin]

Impulse, I_Ball = Δp = m(v' + v)
the answer D is correct!

Why would it be v+v' ?

both v and v' are opposite

 

[tiny-tim]

Why would it be v+v' ?

both v and v' are opposite

the impulse is momentum after minus momentum before …

since v and v' are in opposite directions, that's mv plus mv'

 

[cupid.callin]

OH okay!

so I_Ball = m(v_final - v_initial) = m(v' - (-v) ) = m(v'+v)

I'm being careless again!

Thanks Tiny-tim and ashish

And one more help ...

where does the eqn in hint came from? i mean how do i find it ?

 

[tiny-tim]

where does the eqn in hint came from? i mean how do i find it ?

the rules for impulse are the same as the rules for force

 

[cupid.callin]

So ... the impulse I on ball = Impulse I on block

and thus normal impulse from ground = Impulse of weight + Impulse I

Thus impulse of friction = μ (Impulse of weight + Impulse I)

But this is not the hint

 

[tiny-tim]

impulse of weight = 0

(impulse is over a very short time ∆t …

over that time, impulse of weight = mg∆t, which is infinitesimal compared with the finite impulses of collision)

 

[cupid.callin]

impulse of weight = 0

(impulse is over a very short time ∆t …

over that time, impulse of weight = mg∆t, which is infinitesimal compared with the finite impulses of collision)

I can agree with that but that's not a satisfactory answer

 

[tiny-tim]

I can agree with that but that's not a satisfactory answer

yes it is! ​

check your book on impulse if you don't believe me

(gravity is always left out of impulse equations)