At rest, Block A compresses a spring (k_1 = 200 N/m) by 0.25 m. One the system i

Question

At rest, Block A compresses a spring (k_1 = 200 N/m) by 0.25 m. One the system is released, Block A slides across a surface with a coefficient of kinetic friction of mu_k = 0.2 and strikes Block B which is sitting, at rest, 3 meters from Block A's starting position. After the impact, Block B slides 1 m until it makes contact with another spring (k_2 = 80 N/111) and compresses it 0.2 m before Block B conies to a rest. What is the coefficient of restitution of the impact between Block A and Block B? (Block A weighs 3 N, Block B weighs 1.5 N)

Explanation / Answer

kinetic friction force(Fk)= uk*mg =0.6 ; m-mass of the block A

work done by frictional force(W)= -Fk* d ;d-distance travelled

W= -1.8 J

From Conservation of energy,

change in mechanical energy= W

kx2/2-kx12/2 +0-mv2/2 =W ; When A collides with B, stretched length of the spring = 3-0.25

k*(0.252 -(3-0.25)2) /2 -mv2/2 =-1.8

v= 69.916 ms-1

relative velocity before the collision(Vb)= v= 69.916 ms-1

As before, work done by frictional force(F) ; F= 0.2*mg and distance travelled= 1+0.2

W=0.2*1.2*mg= - 0.36 J

From Conservation of energy,

change in mechanical energy= W

k*(0 -0.22) /2 +m*u2/2-0 = -0.36

velocity of block B just after the collision(u)= 4.025 ms-1

Apply Conservation of momentum for the collision

mA*v= mA*w+ mB*u

velocity of block A (w)= 67.904 ms-1 ; w-velocity of the block A just after the collision with B

Now they are moving just after the collision. Since

e= relative velocity of 2 blocks after collision/relative velocity of 2 blocks before collision

We need to find the relative velocity of A relaive to B

relative velocity after collision(Va)= w-u =63.879

e= Va/Vb =0.914

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