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A ball of clay with a mass of 50.0 g is flying to the right with a speed of 24.0 m/s. A 2.00 kg block is sliding to the left on a frictionless surface with a speed of 1.50 m/s. The ball of clay
collides with the block and sticks to it. What is the magnitude and direction of the velocity of the block and the clay after the collision?

Respuesta :

Answer:

0.878 m/s

Explanation:

mass of ball, m1 = 50 g

initial velocity of ball, u1 = 24 m /s rightwards

mass of block, ,2 = 2 kg

initial velocity of block, u2 = 1.5 m/s leftwards

Let the velocity of both ball and the block after collision is V.

By use of conservation of momentum

Momentum before collision = Momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x V

0.05 x 24 - 2 x 1.5 = (0.05 + 2) x V

1.2 - 3 = 2.05 x V

V = - 0.878 m/s

Thus, both ball and the block will move left after collision with velocity 0.878 m/s.

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