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Consider a one-dimensional collision that involves a body of mass m1 originally moving in the positive x direction with speed v0 colliding with a second body of mass m2 originally at rest. The collision could be completely inelastic, with the two bodies sticking together, completely elastic, or somewhere in between. After the collision, m1 moves with velocity v1 while m2 moves with velocity v2 (a) If m1 > m2, then (A) -v0 (B) 0 (C) 0 (D) v0 (b) and (A) -v0 (B) 0 (C) v/ 2 (D) v0 (c) If m1 (A) -v0 (B) -Vo (C) 0 (D) 0 (d) and (A) -v0 (B) -v0 (C) 0 (D) 0

5 months ago

Answers : (2)

Parth
58 Points
							
See in this case let assume of coefficient of restitution (COR)= e ; -e = (v2-v1)/(u2-u1)
you ‘ve to conserve momentum and energy to find v1,v2, then you ‘ve to compare it with v0 
which become very handy
so if this question is a single correct then assume collision to be perfectly inelastic;
then you’ll find v1 = v2 =m1* v0/(m1+m2)
then for 1 – options
2 – option B
3- option B
4 -option C
coz in option 3 and 4 using the equation of C.O.R ; and conserving momentum you’ll find for m2>m1
v1 can  be negative; v2 can never be negative
hope it helps
pls approve
5 months ago
Vikas TU
12278 Points
5 months ago
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