wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A ball moving translationally collides elastically with another, stationary, ball of the same mass. At the moment of impact the angle between the straight line passing through the centres of the balls and the direction of the initial motion of the striking ball is equal to α=45. Assuming the balls to be smooth, the fraction η of the kinetic energy of the striking ball that turned into potential energy at the moment of the maximum deformation .

Open in App
Solution

If (v1x, v1y) are the instantaneous velocity components of the incident ball and (v2x, v2y) are the velocity components of the struck ball at the same moment, then since there are no external impulsive forces (i.e. other than the mutual interaction of the balls) We have usinα=v1y, v2y=0
mucosα=mv1x+mv2x
The impulsive force of mutual interaction satisfies

ddt(v1x)=Fm=ddt(v2x)

(F is along the x axis as the balls are smooth. Thus Y component of momentum is not transferred.) Since loss of K.E. is stored as deformation energy D, we have

D=12mu212mv2112mv22

=12mu2cos2α12mv1x212mv2x2

=12m[m2u2cos2αm2v1x2(mucosαmv1x)2]

=12m[2m2ucosαv1x2m2v1x2]=m(v1xucosαv1x2)

=m[u2cos2α4(ucosα2v1x)2]

We see that D is maximum when

ucosα2=v1x

and Dmax=mu2cos2α4

Then η=Dmax12mu2=12cos2α=14

On substituting α=45

266505_136193_ans_094311b9d18c4cc3af3a736f7856db1e.png

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Coulomb's Law - Children's Version
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon