Two air carts of mass m1 = 0.85 kg and m2 = 0.45 kg are placed on a frictionless
ID: 1341103 • Letter: T
Question
Two air carts of mass m1 = 0.85 kg and m2 = 0.45 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 = 0.63 m/s . Assume that positive x-axis is directed toward the direction of motion of cart 2.
Part A What is the speed of the two carts at the moment when their speeds are equal? Express your answer using two significant figures.
v = 0.22 m/s <---Correct
Part B How much energy is stored in the spring bumper when the carts have the same speed? Express your answer using two significant figures.
E = 5.8×102 J <---Correct
Part C What is the final speed of the cart 1 after the collision? Express your answer using two significant figures.
v1 =
Part D What is the final speed of the cart 2 after the collision?
v2 =
Explanation / Answer
Given that,
m1 = 0.85 kg ; m2 = 0.45 kg ; K = 690 N/m
v1 = 0 and v2 = 0.63 m/s
A)Let v be the speed of the carts.
From conservation of momentum:
P(i) = P(f)
m1 v1 + m2 v2 = (m1 + m2 )v
0 + 0.45 x 0.63 = (0.85 + 0.45) v
v = 0.2835 / 1.3 = 0.22 m/s
Hence, v = 0.22 m/s
B) Intial energy = KEi = 1/2 m2 v22 = 0.089 J
KEf = 1/2 ( m1 +m2) v2 = 0.031 J
energy stored in the bumper = 0.089 - 0.031 = 0.058 J
Hence, E(stored) = 0.058 J
c)The speed of cart one will be given by:
v1 = 2 m2 v2 / (m1 + m2 ) [derived by using conservation of momentum and energy]
v1 = 2 x 0.45 x 0.63 / (0.85 + 0.45) = 0.44 m/s
Hence, speed of cart 1 = v1 = 0.44 m/s
d)Speed of cart 2 will be:
v2 = ( m1 - m2 ) x v2 / (m1 + m2) = (0.85 - 0.45) x 0.63 / (0.85 + 0.45) = 0.19 m/s
Hence, speed of cart 2 after collision = v2 = 0.19 m/s