A 1.6 kg ball is attached to the end of a 0.40 m string to form a pendulum. This pendulum is released from rest with the string horizontal. At the lowest point of its swing, when it is moving horizontally, the ball collides with a 0.80 kg block that is at rest on a horizontal frictionless surface. The speed of the block just after the collision is 3 m/s. What is the speed of the ball just after the collision

the speed of the ball just after the collision is 1.5 m/s.

Explanation:

Given;

mass of the ball, m₁ = 1.6 kg

initial velocity of the ball, u₁ = 0

mass of the block, m₂ = 0.8 kg

initial velocity of the block, u₂ = 0

final velocity of the block, v₂ = 3 m/s

let the final velocity of the ball after collision = v₁

Apply the principle of conservation of linear momentum for elastic collision;

m₁u₁ + m₂u₂ = m₁v₁  +  m₂v₂

1.6 x 0   +    0.8 x 0       =   1.6 x v₁     +  0.8 x 3

0 = 1.6v₁  + 2.4

-1.6v₁ = 2.4

v₁  = -2.4 / 1.6

v₁ = - 1.5 m/s

v₁ = 1.5 m/s (in opposite direction of the block)

Therefore, the speed of the ball just after the collision is 1.5 m/s.

d is the answer i came up with