A railcar of 2.3 times 10^4 kg of mass moves with a speed of 4.00 m/s. Collides

Question

A railcar of 2.3 times 10^4 kg of mass moves with a speed of 4.00 m/s. Collides and engages two wagons, each of the same mass as the carriage only and move in the same direction with an speed of 2.00 m/s. What is the speed of the three carriages immediately after the collision? How much energy is dissipated in the collision? The body in the image shown is pivoted at O and two forces act on it as shown. If r_1 = 1.30 m, r_2 = 2.15 m: F_1 = 4.20 N, F_2 = 4.90 N, theta_1 = 75.0 degree and theta_2 = 60.0 degree. What is the net torque about the pivot? A rope is wrapped around a solid cylindrical drum (I = 1/2 MR^2). The drum is mounted on a fixed horizontal frictionless axle. The mass of the drum is 30 kg and it has a radius of R = 40.0 cm. The other end of the rope is tied to a block, m = 70.0 kg. Find the acceleration of the block? Assume that the rope does not slip. In the next figure, a thin spherical shell (I = 2/3 MR^2) of radius 10 cm and mass 12 kg starts from rest and rolls without slipping a distance L = 6.0 m down a roof that is inclined at angle theta = 30 degree, the roof's edge is at height H = 5.0 m. What is the angular speed of the sphere about its center as its leaves the roof? Ball A is attached to one end of a rigid massless rod, while an identical ball B is attached to the center of the rod, as shown in the figure. Each ball has a mass of mass m = 0.50 kg, and then of each half of the rod is L = 0.40 m. This arrangement is held by the empty end and is around in a horizontal circle at constant rate, so each ball is in uniform circular motion. Ball A at constant speed v_A = 5.0m/s. Find the tension in each half of the rod

Explanation / Answer

1) a)   speed of three carriages = (2.3 * 104 * 8)/(3 * 2.3 * 104)

                                                    = 2.67 m/sec

    b) energy dissipated = 1/2 * 2.3 * 104 * (16 + 4 + 4 - 3 * 2.672)

                                       =   30052.95 J

2)    Net torque = 4.20 * 1.30 * sin75 - 4.90 * 2.15 * sin60

                           = 3.85 N-m

3) Here, Ma = Mg - T

Also,   T * R = I * alpha

=>    acceleration of block = 2/3 * 9.8

                                        = 6.53 m/sec2

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