A ball of mass m0 is dropped onto a spring-mass-damper system as shown in the fi

Question

A ball of mass m0 is dropped onto a spring-mass-damper system as shown in the figure.

The ball impacts with the block in a perfectly elastic collision, i.e. momentum and kinetic energy are

conserved. Assume the ball is caught after the collision so that it does not strike the block again.

Determine the response of the block x(t) after the collision, given the following values: m = 2 kg ,

b =5 N/(m/s), k =100 N/m,m0 = 0.1 kg , and h = 2 m.

A ball of mass m0 is dropped onto a spring - mass - damper system as shown in the figure. The ball impacts with the block in a perfectly elastic collision, i.e. momentum and kinetic energy are conserved. Assume the ball is caught after the collision so that it does not strike the block again. Determine the response of the block x(t) after the collision, given the following values: m = 2 kg , b = 5 N/(m/s), k = 100 N/m,m0 = 0.1 kg , and h = 2 m.

Explanation / Answer

vo = velocity of ball just before collision = [2gh]^0.5 = 6.26 m/s

let V2 = velocity of block just after collision

then using conservation of momentum and conseravtion of kinetic energy ::

               V2 = 2*mo*Vo/(m0+M) = 0.596 m/s

let at a time instant t , block is displaced by x ,

then balancing force on block ::

   mo*d^x/dt^2   = -kx -bdx/dt

        d^2x/dt^2   = -100x -5dx/dt

      d^2x/dt^2   +100x + 5dx/dt =0

       

solution of 2nd order homogeneuous equation is   ::   x(t) =[A*cos(6.95t)+B*sin(6.95t)] e^-1.25t

apply intial condition , at t=0 , x= 0 , hence    A =0

so    x(t) =B*sin(6.95t)] e^-1.25t

now dx/dt   = B*6.95cos(6.95t)] e^-1.25t -1.25 B*sin(6.95t)] e^-1.25t

given at t=0 , dx/dt = V2 =0.596 ,

hence 0.596 = B*6.95 - 0

                 B =0.085

so   x(t) = B*sin(6.95t)] e^-1.25t = 0.085 *sin(6.95t)] e^-1.25t

i am not able to upload file that is writtent on paper .....

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