A 50 g ball of clay traveling at speed v0 hits and sticks to a 1.0 kg brick sitting at rest on a frictionless surface. Required:
a What is the speed of the block after the collision?
b. What percentage of the ball's initial energy is "lost"?

(a) The speed of the block after the collision is 0.0476v0.

(b) The percentage of the ball's initial energy lost, is 0 % (energy is conserved)

Explanation:

Given;

mass of ball of clay, m₁ = 50 g = 0.05 kg

mass of brick, m₂ = 1 kg

initial velocity of the ball of clay, u₁ = v0

initial velocity of the brick, u₂ = 0

Since the clay ball sticks with the brick after collision, it is inelastic collision.

Therefore, let their final velocity = v

(a) What is the speed of the block after the collision?

Apply the principle of conservation linear momentum

m₁u₁ + m₂u₂ = v (m₁ + m₂)

0.05v₀ + 1(0) = v( 0.05 + 1)

0.05v₀ = 1.05v

v = 0.05v₀ / 1.05

v = 0.0476v₀

Thus, the speed of the block after the collision is 0.0476 of its initial velocity.

(b). What percentage of the ball's initial energy is "lost"?

Initial kinetic energy of the ball = ¹/₂mv₀²

                                                    = ¹/₂ x 0.05 x v₀²

                                                    = 0.025v₀²

Final kinetic energy of the ball, = ¹/₂(m₁ + m₂)v²

                                                    = ¹/₂ x 1.05 x 0.0476v₀²

                                                    = 0.025v₀²

Change in kinetic energy = 0.025v₀² - 0.025v₀²

                                           = 0

percentage change in the initial kinetic energy of the ball;

= (0 / 0.025v₀²) x 100%

= 0 x 100%

= 0 %

Therefore, the percentage of the ball's initial energy lost, is 0 % (energy is conserved)