A 3.0 g coin moving to the right at 27.0 cm/s makes an elastic head-on collision with a 20.0 g coin that is initially at rest. After the collision, the 3.0 g coin moves to the left at 12.5 cm/s.
(b) Find the amount of kinetic energy transferred to the 20.0 g coin.
I found part a, the final velocity of the other coin, to be 5.925 cm/s to the right. I know KE=(1/2)(m)(v^2) but which values do I use for m and v?
To find the amount of kinetic energy transferred to the 20.0 g coin, you can use the equation for kinetic energy: KE = (1/2)mv², where m is the mass of the object and v is its velocity.
In this case, the 20.0 g coin is at rest initially, so its initial kinetic energy is zero. After the collision, the 3.0 g coin moves to the left at 12.5 cm/s and the 20.0 g coin moves to the right at 5.925 cm/s.
To find the amount of kinetic energy transferred to the 20.0 g coin, we need to calculate its final kinetic energy. The final velocity of the 20.0 g coin is calculated as follows:
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
(3.0 g)(27.0 cm/s) + (20.0 g)(0 cm/s) = (3.0 g)(-12.5 cm/s) + (20.0 g)(5.925 cm/s)
Solving the equation, you have:
81 + 0 = -37.5 + 118.5
81 = 81
Therefore, the final velocity of the 20.0 g coin is 5.925 cm/s to the right.
Now that we have the final velocity, we can calculate the final kinetic energy of the 20.0 g coin using the equation:
KE = (1/2)mv²
where m is 20.0 g and v is 5.925 cm/s.
Plug in the values:
KE = (1/2)(20.0 g)(5.925 cm/s)²
Calculating this equation, you will find the amount of kinetic energy transferred to the 20.0 g coin after the collision.
To find the amount of kinetic energy transferred to the 20.0 g coin, we will use the formula for kinetic energy, which is KE = (1/2)mv^2.
Let's assign variables to the given values:
m1 = mass of 3.0 g coin = 3.0 g = 0.003 kg
v1i = initial velocity of the 3.0 g coin = 27.0 cm/s = 0.27 m/s (since it's moving to the right)
v1f = final velocity of the 3.0 g coin = -12.5 cm/s = -0.125 m/s (since it's moving to the left)
m2 = mass of 20.0 g coin = 20.0 g = 0.020 kg
v2f = final velocity of the 20.0 g coin = 5.925 cm/s = 0.05925 m/s (since it's moving to the right)
To find the initial velocity of the 20.0 g coin (v2i), we can use the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision. Mathematically, we can write it as:
(m1 * v1i) + (m2 * v2i) = (m1 * v1f) + (m2 * v2f)
Plugging in the known values, we get:
(0.003 kg * 0.27 m/s) + (0.020 kg * v2i) = (0.003 kg * -0.125 m/s) + (0.020 kg * 0.05925 m/s)
0.00081 kg*m/s + 0.020 kg * v2i = -0.000375 kg*m/s + 0.001185 kg*m/s
Now, let's solve for v2i:
0.020 kg * v2i = -0.000375 kg*m/s + 0.001185 kg*m/s - 0.00081 kg*m/s
0.020 * v2i = -0.00081 kg*m/s + 0.001185 kg*m/s - 0.000375 kg*m/s
0.020 kg * v2i = -0.000
.-
.135 kg*m/s
v2i = (-0.000135 kg*m/s) / 0.020 kg
v2i ≈ -0.00675 m/s (since the 20.0 g coin will have a negative velocity initially)
Now that we have the values for the masses and velocities, we can calculate the amount of kinetic energy transferred to the 20.0 g coin using the kinetic energy formula:
KE = (1/2) * m2 * (v2f)^2
KE = (0.5) * 0.020 kg * (-0.00675 m/s)^2
KE ≈ 0.00009075 J or 9.075 × 10^(-5) J
Therefore, the amount of kinetic energy transferred to the 20.0 g coin is approximately 9.075 × 10^(-5) J.