Question

X0- v 2 w a (1) v=0 As in Figure (a), a 4.04 kg block is attached to a spring with a force constant of 600.00 N/m. The green vertical dashed line indicates the equilibrium position (x= 0). A bullet of 0.05 kg moves toward the block at a speed of vo = 486.00 m/s. The bullet is embedded in the block after the collision as in figure (b). (i) Find the common speed v of the Block + Bullet after the collision. Hint: What physics law is used for collisions? Enter a number Submit (5 attempts remaining) m/s (ii) The Block+Bullet moves on a frictionless horizontal surface and compresses the spring. What is the x in Figure (c) when the Block+Bullet reaches a momentary rest? Report x as NEGATIVE. Enter a number m Submit (5 attempts remaining) (iii) How much time does it take the Block+Bullet to move from figure (b) to figure (c)? Enter a number Submit (5 attempts remaining) (iv) What is the speed of the Block+Bullet when it reaches x, = -0.34 m? Enter a number m/s Submit (5 attempts remaining)

Answer 1

P.PM (6-05) (486)t0 = (005t4.04) PEt KEF Eneagy conpervation, PE, tKe= k xto Ot (405)(5517)- (l00) 22 0.43m (T) Atn) whre T tine puriod 2IT 4.09 T 2m 0S198 t- 0.12 600 : (60)(0.04)-+ (.05)v, U24.28 nl