Two air carts of mass m1=0.84 kg and m2=0.42 kg are placed on a frictionless tra
ID: 1701906 • Letter: T
Question
Two air carts of mass m1=0.84 kg and m2=0.42 kg are placed on a frictionless track. Cart 1 is at rest initially and has a spring bumper with a force constant of 690 N/m. Cart 2 has a flat metal surface for a bumper and moves toward the bumper of the stationary cart with an initial speed v=0.68 m/s. (a)What is the speed of the two carts at the moment when their speeds are equal? (b)How much energy is stored in the spring bumper when the carts have the same speed?(c) What is the final speed of the carts after the collision?Explanation / Answer
A) Applying the conservation of momentum
m2 * v = (m1 + m2) * V
.50 * .68 = 1.30 * V
V = 0.31 m/s
B) Applying the conservation of energy
Energy stored in bumper =
1/2 * m2 * v2 - 1/2 * (m1 + m2) * V2
.50 * .50 * .68 * .68 - .50 * 1.30 * .31 * .31
0.05 J
C) Applying conservation of momentum and energy
m2 * v = m1*v1 + m2 * v2
and
1/2 * m2 * v2 = 1/2 * m1*v12 + 1/2 m2 * v22
From above
m2(v - v2) = m1 * v1
AND
1/2 * m2 * (v2 - v22) = 1/2 * m1*v12
m2 * (v - v2)(v + v2) = m1*v12
Combining
m1 * v1 * (v + v2) = m1*v12
v1 = v + v2
m2 * v = m1*v1 + m2 * (v1 - v)
m2 * v = m1*v1 + m2 * v1 - m2 * v
2 * m2 * v = m1*v1 + m2 * v1
v1 = 2 * m2 * v / (m1 + m2)
v1 = 2 * .50 * .68 / 1.3
v1 = 0.52 m/s