A 0.450-kg ice puck, moving east with a speed of 5.68 m/s , has a head-on collision with a 0.900-kg puck initially at rest. Assume that the collision is perfectly elastic. Part A What is the speed of the 0.450-kg puck after the collision? Express your answer to three significant figures and include the appropriate units. Part B What is the direction of the velocity of the 0.450-kg puck after the collision? Part C What ise the speed of the 0.900-kg puck after the collision? Express your answer to three significant figures and include the appropriate units. Part D What is the direction of the velocity of the 0.900-kg puck after the collision?

Explanation:

Using conservation of linear momentum

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

and since the collision is perfectly elastic

1/2 m₁u₁² + 1/2 m₁u₂² = 1/2 m₁v₁² + 1/2m₂v₂²

head-on collision v₁ = (( m₁ - m₂)/ ( m₁+ m₂)) u₁

and v₂ = ( (2m₁) / (m₁ + m₂) ) u₁

m₁ = 0.450kg, u₁ = 5.68 m/s, m₂ = 0.900kg u₂ = 0 m/s since it is stationary

a) speed v₁ after collision of m₁ =  (( m₁ - m₂)/ ( m₁+ m₂)) u₁ = - 1.89 m/s

b) since the velocity is negative the direction is opposite to the initial direction so it should be west

c) speed v₂ after collision of m₂ = ( (2m₁) / (m₁ + m₂) ) u₁ = 3.79 m/s

d) since it is positive, it should be east

it has to be b, about 97 percent.

i believe the answer is a hope it and is correct