In procedure 1: let\'s asuume that a pair of carts sit on a frictionless track a
ID: 1449586 • Letter: I
Question
In procedure 1: let's asuume that a pair of carts sit on a frictionless track and masses of two carts are different: the small cart has mass m, and the mass of the larger cart is M = 3.46m. Between the carts there is a spring which can "explode", pushing on the two carts at once.
NOTE: Every velocity needs magnitude and direction (given by the sign).
a) Suppose the carts are initially at rest, and after the "explosion" the smaller cart is moving at velocity v = +9.32 m/s.
- Find the velocity of the larger cart. V = _________ m/s
In procedure 2: assume cart 1 has mass m1 = 426 g and cart 2 has mass 129 g. Assume the track is frictionless, and the initial speed of cart 1 is v10 = 4.68 m/s.
If the collision is totally inelastic, find the speed of the carts after the collision.
v = __________ m/s
Explanation / Answer
A pair of carts sit on a frictionless track. Assume the mass of the larger cart is 3.46 times the mass of the?
small cart, In this system, however, there is a spring between the carts which can "explode", pushing on the two carts at once.
let mass of small cart = m
mass of larger cart = 3.46 m
a) Suppose the carts are initially at rest, and after the "explosion" the smaller cart is moving at velocity +9.14 m/s.
Momentum is always conserved!
Initial momentum = 0, since both carts are at rest
Final momentum = (m * 9.32) + (3.46 * m * v)
Final momentum = Initial momentum
(m * 9.32) + (3.46 * m * v) = 0
(m * 9.32) = -(3.46 * m * v)
divide both sides by m
9.32 = -(3.46 * v)
v = 9.32 / -3.46 = - 2.69 m/s
The velocity of the larger cart = 2.69 m/s in the opposite direction of smaller car!
b)
m1v1 + m2v2 = (m1 + m2) V
=> V = 0.426(4.68) / (0.426 + 0.129) + 0 / (0.426 + 0.129)
=> V = 3.59 m/s