URGENT HELP PHYSICS CART

Question

In the physics lab, a cart of mass m₁ = 0.30 kg moves with velocity v₁ = 2.5 m/s on a frictionless track. It collides with another cart of mass m₂ = 0.40 kg, which is initially stationary. The carts stick together and roll off of the end of the track onto the rough surface of the table. The carts roll across the table for 0.3 seconds before they roll off the edge of the 1.0-meter-tall table.

(A) What kind of collision occurs between the carts?

(B) What is the velocity of the carts immediately after the collision?

(C) The carts (still stuck together) land 20 cm away from the bottom of the table. What was the velocity of the carts when they fell off the table?

(D) What is the coefficient of kinetic friction between the table and the carts?

Answer 1

A) This is an inelastic collision.

B) The total momentum of the system before the collision, including both carts is contained in the moving cart (because a stationary cart has 0 momentum). That mv = (0.3 kg)(2.5 m/s) = 0.75 kgm/s. The total momentum after the collision is the same, but now the carts are stuck together, so we add the masses together and solve for the new velocity: 0.75 = (0.3kg + 0.4kg)v_new ; v_new = 0.75 / 0.7) = 1.07 m/s

C) We first calculate how long it takes for the carts to fall 1 m: use h_f = h_i + v_vot + 1/2gt² with g = -9.9 m/s/s and v_vo = 0; -1 = 0 + 0 - 4.9t²; t = sqrt (1 / 4.9) = 0.452 s

Horizontal v is constant during the fall and it moved 0.2 m horizontally,

therefore v = d/t = 0.2 / 0.452 = 0.44 m/s

D) Velocity dropped from 1.07 m/s to 0.44 m/s in o.3 s, therefore a = (0.44 - 1.07) / 0.3 = -2.1 m/s/s; F = ma = (0.7)(-2.1) = -0.147 N this represents the force of friction F_f; coefficient of friction μ = F_f / normal force F_N and normal force is equal to weight because this is flat; weight = mg = (0.7)(9.8) = 6.86 N,

so μ = F_f / F_N = 0.147 / 6.86 = 0.02.