As in procedure 1: a pair of carts sit on a frictionless track. Assume the mass
ID: 2097218 • Letter: A
Question
As in procedure 1: a pair of carts sit on a frictionless track. Assume the mass of the larger cart is 3.65 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..
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 +3.5 m/s.
- Find the velocity of the larger cart.
- If the mass of the smaller cart is 4.05 kg, find the energy supplied by the spring to the carts.
- If the spring has spring constant k = 695 N/m, find the distance the spring was compressed before the "explosion".
HINT: Think conservation of energy!
b) Suppose the carts are initially moving together at constant velocity +1.37 m/s. After the "explosion", the smaller cart is moving at velocity +3.5 m/s. Find the velocity of the larger cart.
In procedure 2: assume the mass of cart 1 is m1 = 234 g and the mass of cart 2 is 181 g. Assume the track is frictionless, and the initial speed of cart 1 is 1.85 m/s.
a) If the collision is totally inelastic, find V, the speed of the carts after the collision.
b) Calculate the energy lost in the collision.
Reminder: Speed is the magnitude of the velocity vector.
In procedure 3: assume cart 1 has mass 409 g and cart 2 has mass 627 g. Assume the track is frictionless, and the initial speed of cart 1 is 2.63 m/s. Find:
- v1, the speed of cart 1 after the collision
- v2, the speed of cart 2 after the collision
Explanation / Answer
A pair of carts sit on a frictionless track. Assume the mass of the larger cart is 3.48 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..
NOTE: Every velocity needs magnitude and direction (given by the sign).
let mass of small cart = m
mass of larger cart = 3.48 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 * 9.14) + (3.48 * m * v)
Final momentum = Initial momentum
(m * 9.14) + (3.48 * m * v) = 0
(m * 9.14) = -(3.48 * m * v)
divide both sides by m
9.14 = -(3.48 * v)
v = 9.14
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.