Question

Consider that you have two blocks and they are connected to each other with a spring. Block AA  has mass 1.00 kgkg, and block BB has mass 3.00 kgkg. The blocks are compressed with a spring SS between them; then the system is released from rest on a level, frictionless surface. The spring, which has negligible mass, is not fastened to either block and drops to the surface after it has expanded. The spring has force constant 720 N/mN/m and is initially compressed 0.225 mm from its original length.

A)Find the acceleration of block A just after the blocks are released.

B)Find the final velocity of block A after the blocks leave the spring?

C)Find the final velocity of block B after the blocks leave the spring?

Answer 1

pls confirm compression distance =0.225mm= 0.255x10^-3m. i am doing here this values

a) force acting on both masses will same

force acting on mass A = F=M_axa_a

force acting by spring= =kx

k= spring constant x= compression distance   a_a=acceleration of mass A

acceleration of mass A = a_a =kx/_Ma =720x0.255x10^-3/_1kg =0.162m/s²

b) apply momentum conservation at initial position and final position of the system

initioal momentumPi =0 ........./because velocity of both masses is zero

Va= speed of mass A

Vb=speed of mass B

final momentum Pf =Ma Va - Mb Vb ..../ negative comes by veloity direction are oppposte each other

Pi = f

0=Ma Va - Mb Vb

Va =^{(Mb Vb)}/_Ma= 3Vb/1 =3Vb

Vb=Va/3

apply energy conservation

KEi=0

PEi=  ¹/₂kx²

KEf=¹/₂ Ma Va²+¹/₂ Mb Vb²

PEf =0

¹/₂kx² =¹/₂ Ma Va²+¹/₂ Mb Vb² .............../put Vb=Va/3

kx² = Ma Va²+Mb (Va/3)²

k=720N/m

x=0.225m

720x(0.255x10^-3)² =Va²(1+¹/3)

Va =5.228x10^-3m/s

c)

Vb=Vb=Va/3 =(5.228x10^-3m/s)/3 =1.748x10^-3