Could someone me with this question? an answer and a detailed explanation would be appreciated. in advance!

question is in the image

a) 0.20 m/s

b) 0.40 m/s

c) 0.12 m/s

Explanation:

Momentum before collision = momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

a) After the collision, they stick together and move at the same speed, so v₁ = v₂.

m₁u₁ + m₂u₂ = (m₁ + m₂) v

(0.20 kg) (0.60 m/s) + (0.40 kg) (0 m/s) = (0.20 kg + 0.40 kg) v

0.12 kg m/s = 0.60 kg v

v = 0.20 m/s

b) If v₂ = 0.10 m/s:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.20 kg) (0.60 m/s) + (0.40 kg) (0 m/s) = (0.20 kg) v + (0.40 kg) (0.10 m/s)

0.12 kg m/s = 0.20 kg v + 0.04 m/s

0.08 kg m/s = 0.20 kg v

v = 0.40 m/s

c) If v₁ = v₂ / 2:

(0.20 kg) (0.60 m/s) + (0.40 kg) (0 m/s) = (0.20 kg) v₁ + (0.40 kg) (2v₁)

0.12 kg m/s = 0.20 kg v₁ + 0.80 kg v₁

0.12 kg m/s = 1.00 kg v₁

v₁ = 0.12 m/s

the answer is d. 4

hope this

also the more points you give the faster your questions get answered

current flows in a wire