A 4.0 g coin sliding to the right at 26.0 cm/s makes an elastic head-on collision with a 12.0 g coin that is initially at rest. After the collision, the 4.0 g coin moves to the left at 13 cm/s.

(a) Find the final velocity of the other coin.
cm/s
(b) Find the amount of kinetic energy transferred to the 12.0 g coin.
J

To solve this problem, we can use the principles of conservation of momentum and conservation of kinetic energy in an elastic collision.

(a) Find the final velocity of the other coin:

1. Start by writing down the given information:
- Mass of the first coin (4.0 g) = 0.004 kg (since 1 g = 0.001 kg)
- Initial velocity of the first coin (v₁) = 26.0 cm/s = 0.26 m/s (since 1 cm/s = 0.01 m/s)
- Mass of the second coin (12.0 g) = 0.012 kg
- Final velocity of the first coin (v₂) = -13 cm/s = -0.13 m/s

2. Apply the conservation of momentum equation:
The total momentum before the collision is equal to the total momentum after the collision.
(m₁ * v₁) + (m₂ * v₂) = (m₁ * v₁') + (m₂ * v₂')

Plugging in the given values:
(0.004 kg * 0.26 m/s) + (0.012 kg * 0) = (0.004 kg * v₁') + (0.012 kg * v₂')

Simplifying:
0.00104 kg∙m/s = 0.004 kg * v₁'
v₁' = 0.00104 kg∙m/s / 0.004 kg
v₁' ≈ 0.26 m/s

The final velocity of the other coin is approximately 0.26 m/s to the right.

(b) Find the amount of kinetic energy transferred to the 12.0 g coin:

1. Calculate the initial kinetic energy of the system:
Kinetic energy before collision = (1/2) * (m₁ * v₁^2) + (1/2) * (m₂ * 0^2)
= (1/2) * (0.004 kg * (0.26 m/s)^2) + (1/2) * (0.012 kg * 0^2)

Simplifying:
= (1/2) * (0.004 kg * 0.0676 m^2/s^2) + 0

= 1.35 x 10^-4 J

2. Calculate the final kinetic energy of the system:
Kinetic energy after collision = (1/2) * (m₁ * v₁'^2) + (1/2) * (m₂ * v₂'^2)
= (1/2) * (0.004 kg * (0.26 m/s)^2) + (1/2) * (0.012 kg * (0.13 m/s)^2)

Simplifying:
= (1/2) * (0.004 kg * 0.0676 m^2/s^2) + (1/2) * (0.012 kg * 0.0211 m^2/s^2)

= 6.76 x 10^-5 J + 1.27 x 10^-5 J

= 8.03 x 10^-5 J

3. Determine the amount of kinetic energy transferred:
The kinetic energy transferred is equal to the difference between the initial and final kinetic energy:
Kinetic energy transferred = Kinetic energy before collision - Kinetic energy after collision
= 1.35 x 10^-4 J - 8.03 x 10^-5 J
= 5.49 x 10^-5 J

The amount of kinetic energy transferred to the 12.0 g coin is approximately 5.49 x 10^-5 J.

m₁=0.004 kg, v₁₀=0.26 m/s

m₂=0.012 g, v₂= -0.13 m/s
v₁=?

m₁v₁₀ = m₁v₁ +m₂v₂

v₁= (m₁-m₂)v₁₀/(m₁+m₂)

ΔKE = m₁v₁₀²/2- m₁v₁²/2=
=m ₂v₂²/2= …