Question

2. | 10 The block B of mass m3 = 2.5 kg is attached to a spring with spring constant k, and is initially at rest at point A as shown in the diagram. The block is then released and slides inside the curved slot until it collides at location C with a ball S that is hanging by a rope from a fixed location D. After the collision, block B is at rest at location C, and the ball swings on the rope up to a maximum angle a at location E. The ball has mass ms = 12 kg. The spring's unstretched length is 1.5 m. The slot is frictionless, in the shape of a thin quarter circle of radius 2 meters. The block and ball are both very small. (a) Assume k = 100 N/m. Calculate the force of the slot on the block at point A. Also, if block B has a speed of Vc at the instant before g=9.81 m/s? it hits the ball at point C, calculate -2.0 m- the force of the slot on B at that time in AB X-y coordinates, in terms of Vc. (Do not try to determine a numerical value for V.). 10 (b) Now assume k = 130 N/m. Calculate the velocity of the block Vc at the instant before it hits the ball at point C. 2.0 m 1.0 m 107 (c) If the rebound angle of the ball's rope after the collision is found to be a = 30 degrees, what is the spring constant k in this case? 30 marks for question #2

Answer 1

mas 2m het us generalise the Result using point mass for simplicity Given Af: 2.0m DC=im AG=FG=2m FD=2-1=lm unstreched spring length=1.5m AN = AFP+ FA? AD² = 2² + 1² AD = 2.236 m two - FDŹ 0 = 26.565° ca At position A spring force a ka = K (AD -1.5) =100 (2-236-05) iii force exdited by slat at j NEE e position A (N) = kxloso.. NFE 73.6 x cos 26-565 IN, 65.03.] Aus- (as AF 73.6 N Ku

apply conservation of enugy b/w points A and KEA & PEA = K.Ect P. Ec o+mgxha - I m Vc² to 2.5 x 9.81 x2 = 4x2.5 X Ve het the velocity at point cin ve het fore applied by stact at Cis No N-kx=mve² Ne=kx+mve? Nc = loo (1.5-1) + 2-5 Ve? I Nc= 50 +1. 25V,²l Applying consuvation of enugy blog paint A and c b Joho К. Ее + Р. Є. . Е. Р. . o + mg hou tot kx² = 1 mive²t 2 k to 2.5x9.01x 2 + 1 K (2136-1-5) = m VE +K (1-5-1) 4 9.05 + 0 - 21k : 1 2 3 v 0 125k 149.05 +0.145k a 1.25ve?

liven k= 130 N/m 49.05+0.145x130 = 1.25 VC? lvc=7.37mis sus-b 1 (C) het the velocity of block B before - collision is vc and afteu colission velocity of balls in vs Applying consuvation of momentuan of Mc Vc = ms sto { beo block B rests . after collisions - 2.50VC = 12 Vs Vs = 2.5 Vc 12 wen Applying consavation of enugy between cand E aftu collision Ź x12.0 Vse KES+ . Es K.6E+P.EE Im vs to = 0 + mgh -o savs h = DS - DE cosa hi 1-1C 30° - ha 0.13am using eq Ix120x Vs? = 12 x 9.81 x 0.1334 [Vs= 1.621 m/s Vs = 2.5 Vc Ve = 7.782 is using eqa o 49.05 + Oilus k = 1:25 Vc? 49.05 + 0.145 k = 1.25 X 7.78? i [K= 183.014 N/m] sus (6) 12 d