## Wednesday, October 20, 2010

### Multiple Choice Questions on Elastic and Inelastic Collisions

Today we will discuss two questions on collisions. The first question pertains to elastic collision in one dimension and the second question pertains to inelastic collision in two dimensions:

(1) A particle A of mass m moving along the positive x-direction with kinetic energy K suffers an elastic head-on collision with a stationary particle B of mass M. After the collision the particle A moves along the negative x-direction with kinetic energy K/9. What is the mass of particle B?

(a) 9 m

(b) 6 m

(c) 3 m

(d) 2 m

(e) m/3

Since the kinetic energy of A after collision is one-nineth of its initial kinetic energy, the momentum of A after collision is one-third of its initial momentum.

Since the momentum is to be conserved, we have

p = p p/3 where p is initial momentum of A and p’ is the momentum of B after the collision.

[The final momentum of particle A is negative since its direction is reversed].

Therefore, p’ = 4p/3

The kinetic energy gained by particle B due to the collision is p2/2M where M is the mass of particle B.

The kinetic energy lost by particle A due to the collision is (8/9)×p2/2m.

[Note that the initial kinetic energy of particle A is p2/2m and its final kinetic energy is (1/9) p2/2m].

Since the kinetic energy too is conserved in elastic collisions, the kinetic energy gained by particle B is equal to the kinetic energy lost by particle A. Therefore, we have

p2/2M = (8/9) p2/2m

Substituting for p’ = 4p/3, we have

(16/9) (p2/2M) = (8/9) p2/2m from which M = 2 m.

(2) A particle of mass m moving westwards with speed v collides with an identical particle moving northwards with speed v. If they stick together after the collision, what is their common speed after the collision?

(a) 4v

(b) 2√2 v

(c) 2v

(d) √2 v

(e) v/√2

The total momentum before collision is the vector sum of the initial momenta mv and mv at right angles (since one is westwards and the other is northwards) and is equal to √2 mv acting along the north-west direction (Fig.).

The combined mass is 2m. If V is the common speed, the total momentum after collision is 2m V, acting along the north-west direction.

Since the momentum is conserved, we have

2m V = √2 mv so that V = v/√2

1. beautiful job sir ji..........
sir,aapne blogspot me mathmatical term i.e.squar,under root kaise likha please tell me

2. Hello Prashanth,
I can't make out what you want. I thought you were asking about superscripts and subscripts. But I found them in your blog!

3. Anonymous5:54 PM

I was wondering if you ever considered changing the structure of your site?
Its very well written; I love what youve got to say.

But maybe you could a little more in the way of content so people could connect
with it better. Youve got an awful lot of text for only having one or
2 images. Maybe you could space it out better?
Also visit my web site :: seo schweiz

4. Anonymous7:19 PM

Excellent post. I was checking continuously thhis blog annd I am impressed!
Extremely helpful info particularly the last part :
) I care for such information a lot. I was seeking this particular information ffor a very long time.
Thank you and best of luck.

Feel free too visiut my site ... jp Morgan chase new york address

5. Anonymous10:00 AM

I must thank you for the efforts you've put inn writing this website.
I'm hoping to check out the same high-grade congent from you later on as well.

In truth, your creative writing abilities has inspired me to get my own, persnal webbsite now ;)

Also visit my blog post internet Banking in India report