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Blocks A (mass 3.00 kg ) and B (mass 9.00 kg ) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 5.00 m/s . The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line. Let +x be the direction of the initial motion of block A. Find the velocity of block B when the energy stored in the spring bumpers is maximum.

Respuesta :

Answer:

The velocity of block B when the energy stored in the spring bumpers is max = 2.5m/s

Explanation:

An head on collision is aka elastic collision and in elastic collision momentum and energy is conserved that is the total momentum before collision = total momentum after collision

Given

Mass of block A mA = 3kg, initial Velocity of block A vA1 = 5m/s, Final velocity = vA2

Mass of block B mB = 9kg, Initial velocity of block B vB1 = 0m/s, Final velocity = vB2

Therefore we have the equation

1)mAvA1 + mBvB1 = mAvA2 + mBvB2

Also in elastic collision between 2 objects, the relative velocities before and after the collision have the same magnitude but opposite direction

vA1 - vB1 = vB2 - vA2

vA2 = vB2 - vA1 when vB1 = 0

Substitute into the equation we have

(3*5)+(9*0) = 3*(vB2-vA1) + (9*vB2)

15 = 3vB2 - 3vA1 +9vB2

15=3vB2 - (3*5) + 9vB2

15+15 = 12vB2

vB2 = 2.5m/s

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