Two billiard balls are initially traveling toward each other at speeds of 1.15 m
ID: 1476173 • Letter: T
Question
Two billiard balls are initially traveling toward each other at speeds of 1.15 m/s for ball 1 and 4.60 m/s for ball 2. The balls undergo an elastic, head-on collision. Find their final velocities.
ball 1
ball 2
A particle with an initial linear momentum of 5.02 kg · m/s directed along the positive x-axis collides with a second particle, which has an initial linear momentum of 10.04 kg · m/s, directed along the positive y-axis. The final momentum of the first particle is 7.53 kg · m/s, directed 45.0° above the positive x-axis. Find the final momentum of the second particle.
magnitude direction ---Select--- in the same direction it was initially going. opposite the direction it was initially going.Explanation / Answer
here,
mass of ball 1, m1 = m
mass of ball 2, m2 = m
Velocity of ball 1 , v1 = 1.5 m/s
Velocity of ball 2 , v2 = -4.60 m/s
in an head on collision we ahve :
Final Velocity of Ball 2 :
V2f= (2m1v1)/(m1+m2) + ((m2 - m1)*V2) / (m1+m2)
V2f = (2m*1.5)/(m+m) + 0 ( as m1 = m2 = m)
V2f = 1.5 m/s
Therefore
From conservation of momentum :
before collision = after collision
m1v1 + m2v2 = m1v1f + m2v2f
v1 + v2 = v1f + v2f
Solving for final velocity of ball 1, V1f
v1f = v1 + v2 - v2f
V1f = 1.5 - 1.5 - 4.60
V1f = -4.65 m/s
Final Velocity are :
V1f = -4.65 m/s
V2f = 1.5 m/s
(When mass are same velcoity get's interchanged between particles)
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
PART B:
intial momentum of particle 1 , P1x = 5.02 kg.m/s
intial momentum of particle 2 , P2y = 10.04 kg.m/s
Final momentum of particle 1 ,
P1'x = 7.53*Cos45
P1'y = 7.53*Sin45
From conservation of momentum : initial = final
in x axis,
P1x + p2x = p1'x + p2'x
5.02 + 0 = 7.53*Cos45 + p2'x
p2'x = 5.02 - 7.53*cos45
p2'x = -0.305 kg.m/s
Similarly in y axis::
P1y + p2y = p1'y + p2'y
0 + 10.04 = 7.53*Sin45 + p2'y
p2'y = 10.04 - (7.53*Sin45)
p2'y = 4.715 kg.m/s
Net momentum will be :
Pnet = sqrt(P2'x^2 + P2'y^2)
Pnet = sqrt ( (0.305)^2 + 4.715^2 )
Pnet = 4.725 kg.m/s
Direction can be found out by :
A = arcTan(y/x)
A = arcTan(4.715/0.305)
A = 86.3 degrees towards west or negative x axis