What is the speed of each marble immediately after the collision?
October 22nd, 2009 | by admin |A 46g marble moving at 2.3m/s strikes a 23g marble at rest. What is the speed of each marble immediately after the collision?
Apply using the following theory:
Assume an elastic collision:
initial KE = 1/2 m1 v1^2
=
final KE = 1/2 m1 v1f^2 + 1/2 m2 v2f^2
initial momentum = m1 v1
=
final momentum = m1 v1f + m2 v2f
Two equations, two unknowns
Solve the second equation for the second marble’s speed after collision:
v2f = m1 (v1 – v1f) / m2
Plug that into the first:
m2 v1^2 = m2 v1f^2 + m1 (v1 – v1f)^2
0 = v1f^2 (m2 + m1) – 2 m1 v1 vf – v1^2 (m2 – m1)
And solve the quadratic for v1f
v1f = 2 m1 v1 +- sqrt (4m1^2 v1^2 + 4 v1^2 (m2^2 – m1^2) ) / 2 (m2+m1)
= v1 (m1 +- m2) / (m2 + m1)
= v1 or v1 (m1 – m2) / (m1 + m2)
The first solution means it just went through without touching the other marble. The second one is the one you want.
And substitute to get the final speed of the second ball:
v2f = m1 (v1 – v1f) / m2
= 2 m1 v1 / (m1 + m2)
One Response to “What is the speed of each marble immediately after the collision?”
By Ananda on Oct 22, 2009 | Reply
Apply using the following theory:
Assume an elastic collision:
initial KE = 1/2 m1 v1^2
=
final KE = 1/2 m1 v1f^2 + 1/2 m2 v2f^2
initial momentum = m1 v1
=
final momentum = m1 v1f + m2 v2f
Two equations, two unknowns
Solve the second equation for the second marble’s speed after collision:
v2f = m1 (v1 – v1f) / m2
Plug that into the first:
m2 v1^2 = m2 v1f^2 + m1 (v1 – v1f)^2
0 = v1f^2 (m2 + m1) – 2 m1 v1 vf – v1^2 (m2 – m1)
And solve the quadratic for v1f
v1f = 2 m1 v1 +- sqrt (4m1^2 v1^2 + 4 v1^2 (m2^2 – m1^2) ) / 2 (m2+m1)
= v1 (m1 +- m2) / (m2 + m1)
= v1 or v1 (m1 – m2) / (m1 + m2)
The first solution means it just went through without touching the other marble. The second one is the one you want.
And substitute to get the final speed of the second ball:
v2f = m1 (v1 – v1f) / m2
= 2 m1 v1 / (m1 + m2)
References :
(Ω)Mistress Bekki