In the solution for the following problem the conservation ofmomentum equation i
ID: 1671478 • Letter: I
Question
In the solution for the following problem the conservation ofmomentum equation is used m1v1+m2v2=m1v1'+m2v2' I was wondering if you could use v1'=(m1-m2/m1+m2)v1 and v2'=(2m1/m1+m2)v1 If you cannot, I was wondering why? A .060 kg tennis ball, moving with a speed of 2.5m/s, collideshead-on with a .090kg ball initially moving away from it at a speedof 1.15m/s. Assuming a perfectly elastic collision, what arethe speed and direction of each ball after the collision? Thanks In the solution for the following problem the conservation ofmomentum equation is used m1v1+m2v2=m1v1'+m2v2' I was wondering if you could use v1'=(m1-m2/m1+m2)v1 and v2'=(2m1/m1+m2)v1 If you cannot, I was wondering why? A .060 kg tennis ball, moving with a speed of 2.5m/s, collideshead-on with a .090kg ball initially moving away from it at a speedof 1.15m/s. Assuming a perfectly elastic collision, what arethe speed and direction of each ball after the collision? ThanksExplanation / Answer
Masses m = 0.06 kg
M = 0.09 kg
Initial speeds u = 2.5 m / s
U = 1.15 m / s
From law of conservation of momentum , mu+MU = mv +MV
0.15 + 0.1035 = 0.06 v + 0.09V ---( 1)
For perfectly elastic collision ,coefficient ofrestitution e = 1
( V –v ) / ( u-U ) = 1
V –v = u –U
= 1.35
V = v+ 1.35 --( 2)
Plug eq ( 2 ) in eq ( 1) we get ,
0.2535 = 0.06 v + 0.09 ( v + 1.35 )
=0.06 v + 0.09 v + 0.1215
= 0.15 v + 0.1215
v = 0.88 m / s
from eq ( 2) V = 2.23 m / s
Both the balls moves in initial direction