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A track consists spring launcher on one end. A spring which is compressed 0.5 m has a
spring constant of 25 N/m. A 0.05 kg ball is resting at the end of the spring. When
released, the spring pushes the ball down the track. It collides and sticks to a second
block of 0.05 kg at rest.

Respuesta :

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(a) The velocity of the first ball before the collision with the second ball is 11.18 m/s.

(b) The final velocity of the two balls after the collision is determined as 5.59 m/s.

Speed of the block when pushed by the spring

The speed of the block when pushed by the spring is calculated as follows;

K.E = Ux

¹/₂mv² = ¹/₂kx²

mv² = kx²

v² = kx²/m

v² = (25 x 0.5²)/0.05

v² = 125

v = 11.18 m/s

Final velocity of the two balls after the collision

The velocity of the two balls after the collision is calculated as follows;

Pi = Pf

where;

  • Pi is initial momentum
  • Pf is final momentum

m1u1 + m2u2 = v(m1 + m2)

0.05(11.18) + 0.05(0) = v(0.05 + 0.05)

0.559 = 0.1v

v = 5.59 m/s

Learn more about velocity here: https://brainly.com/question/4931057

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