Blocks A (mass 8.00 kg ) and B (mass 12.00 kg , to the right of A) move on a fri

Question

Blocks A (mass 8.00 kg ) and B (mass 12.00 kg , to the right of A) move on a frictionless, horizontal surface. Initially, block B is moving to the left at 0.500 m/s and block A is moving to the right at 2.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is headon, so all motion before and after it is along a straight line. Let +x be the direction of the initial motion of A

Find the maximum energy stored in the spring bumpers.

Find the velocity of block A when the energy stored in the spring bumpers is maximum.

Find the velocity of block B when the energy stored in the spring bumpers is maximum.

Find the velocity of block A after the blocks have moved apart.

Find the velocity of block B after the blocks have moved apart.

Express your answer with the appropriate units.

Explanation / Answer

m1 (A) = 8 kg   m2 (B) = 12 kg

speeds before collision

u1 = 2 m/s     u2 = -5 m/s

speeds after collision

v1 (A) = ?        v2 (B) = ?

1)
Umax = 0.5*m1*u1^2 + 0.5*m2*u2^2

Umax = (0.5*8*2^2)+(0.5*12*0.5^2) = 17.5 J

(2)

vA = 0

(3)

vB = 0

(4)

initial momentum before collision

Pi = m1*u1 + m2*u2

after collision final momentum

Pf = m1*v1 + m2*v2

from moentum conservation

total momentum is conserved

Pf = Pi

m1*u1 + m2*u2 = m1*v1 + m2*v2

(8*2) - (12*0.5) = (8*v1) + (12*v2).....(1)

from energy conservation

total kinetic energy before collision = total kinetic energy after collision

KEi = 0.5*m1*u1^2 + 0.5*m2*u2^2

KEf =   0.5*m1*v1^2 + 0.5*m2*v2^2

KEi = KEf

0.5*m1*u1^2 + 0.5*m2*u2^2 = 0.5*m1*v1^2 + 0.5*m2*v2^2

(0.5*8*2^2)+(0.5*12*0.5^2) = 0.5*8*v1^2 + 0.5*12*v2^2 ........(2)

v1 = -1 m/s <---------------answer

v2 = +1.5 m/s <<---------answer

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