6.29 Which of the following potential energy curves in Fig. 6.18 cannot possibly describe the elastic collision of two billiard balls ? Here r is the distance between centers of the balls.

0 2 Views | Posted 4 months ago
Asked by Shiksha User

  • 1 Answer

  • P

    Answered by

    Payal Gupta | Contributor-Level 10

    4 months ago

    6.29 The potential energy of two masses in a system is inversely proportional to the distance between them. The potential energy of the system of two balls will decrease as they get closer to each other. When the balls touch each other, the potential energy becomes zero, I.e. at r = 2R. The potential energy curve in (i), (ii), (iii), (iv) and (vi) do not satisfy these conditions. So there is no elastic collision.

Similar Questions for you

V
Vishal Baghel

Using Newton’s formula,

( f + d 1 ) ( f d 2 ) = f 2

f = d 1 d 2 d 1 d 2 R = 2 d 1 d 2 d 1 d 2

P
Payal Gupta

This is a multiple choice answer as classified in NCERT Exemplar

(c) When M1 comes in contact with the spring. M1 is retarded by the spring force and M2 is accelerated by the spring force. 
 
a) So spring will continue compress until the blocks moves with the same velocity. 
 
b) As the surfaces are frictionless momentum of the system will conserved. 
 
c) If the spring is massless whole energy of M1 will be imparted and M1 will be rest . 
 
d) collision is elastic even if friction is not involved.

P
Payal Gupta

This is a multiple choice answer as classified in NCERT Exemplar

(b) conserving energy between “O” ans ”A”

Ui + Ki = Uf + Kf

0+1/2mv2= mgh + 1/2mv’

( v ' ) 2 2 = v 2 2 = - g h

(v’)2=v2-2gh = v= v 2 - 2 g h ……….1

Let speed after emerging be v1 then

=1/2mv12=1/2[1/2mv’2]

1/2m(v1)2=1/4m(v’)2=1/4m[v2-2gh]

V1= v 2 2 - g h ………….2

From eqn 1 and 2

v ' v 1 = v 2 - 2 g h v 2 - 2 g h / 2 = 2

So v1 = v’/ 2 =v2(v’/2)

v1>v’/2

hence after emerging from the target velocity of the bullet is more than half of its earlier velocity v’

(d) as the velocity of the bullet changes to v’ which is less than v’ hence , path

...more
P
Payal Gupta

This is a multiple choice answer as classified in NCERT Exemplar

(b, d) When a man of mass m climbs up the staircases of height L, work done by the gravitational force on the man is mgl work done by internal muscular forces will be mgL as the change in kinetic is almost zero.

Hence total work done =-mgL + mgL=0

As the point of application of the contact forces does not move hence work done by reaction forces will be zero.

P
Payal Gupta

This is a multiple choice answer as classified in NCERT Exemplar

(c) m =150g =3/20kg

Time of contact =0.001s

U=126km/h= 126 × 1000 60 × 60 = 35 m s

V= -35m/s

Change in momentum of the ball = m (v-u)= 3 20 - 35 - 35 k g m / s

=21/2

F= dp/dt=- 21 / 2 0.001 N = - 1.05 × 10 4 N

Here – negative sign indicates that force will be opposite to the direction of movement of the ball before hitting.

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 688k Reviews
  • 1800k Answers

Learn more about...

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

×
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.

Need guidance on career and education? Ask our experts

Characters 0/140

The Answer must contain atleast 20 characters.

Add more details

Characters 0/300

The Answer must contain atleast 20 characters.

Keep it short & simple. Type complete word. Avoid abusive language. Next

Your Question

Edit

Add relevant tags to get quick responses. Cancel Post