Sarthaks Test
0 votes
in Physics by (8.6k points)

A ball of mass m, moving with a speed 2v0, collides inelastically (e > 0) with an identical ball at rest. Show that 

(a) For head-on collision, both the balls move forward. 

(b) For a general collision, the angle between the two velocities of scattered balls is less than 90°.

1 Answer

0 votes
by (13.6k points)
selected by
Best answer

(a) For head on collission:

Conservation of momentum ⇒ 2mv0 = mv1 + mv2 

Since e <1 ⇒ v1 has the same sign as v0, therefore the ball moves on after collission.

(b) Conservation of momentum ⇒  p = p1+ p2

⇒ p2/2m > p22/2m + p22 

Thus p, p1 and p2 are related as shown in the figure.

θ is acute (less than 90°) (p2 = p12 + p22 would give θ = 90°)

Related questions

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.