A 5.9 g coin sliding to the right at 23.4 cm/s makes an elastic head-on collision with a 17.7 g coin that is initially at rest. After the collision, the 5.9 g coin moves to the left at 11.7 cm/s.Find the amount of kinetic energy transferred to the 17.7 g coin.
To find the amount of kinetic energy transferred to the 17.7 g coin during the elastic collision, we can use the principle of conservation of kinetic energy.
The principle of conservation of kinetic energy states that the total kinetic energy before the collision is equal to the total kinetic energy after the collision.
Kinetic energy is given by the formula: KE = 1/2 * m * v^2
Where:
KE = Kinetic energy
m = Mass of the object
v = Velocity of the object
Let's denote the 5.9 g coin as Coin 1 and the 17.7 g coin as Coin 2.
Before the collision:
Coin 1:
Mass (m1) = 5.9 g = 0.0059 kg (converting grams to kilograms)
Velocity (v1) = 23.4 cm/s = 0.234 m/s (converting cm/s to m/s)
Coin 2:
Mass (m2) = 17.7 g = 0.0177 kg (converting grams to kilograms)
Velocity (v2) = 0 m/s (initially at rest)
The total kinetic energy before the collision is:
KE_initial = KE(Coin 1) + KE(Coin 2)
= (1/2 * m1 * v1^2) + (1/2 * m2 * v2^2)
= (1/2 * 0.0059 kg * (0.234 m/s)^2) + (1/2 * 0.0177 kg * (0 m/s)^2)
= 0.00726 J
After the collision:
Coin 1:
Mass (m1) = 5.9 g = 0.0059 kg
Velocity (v1) = -11.7 cm/s = -0.117 m/s (moving to the left, so negative velocity)
Coin 2:
Mass (m2) = 17.7 g = 0.0177 kg
Velocity (v2) = ? (unknown)
The total kinetic energy after the collision is:
KE_final = KE(Coin 1) + KE(Coin 2)
= (1/2 * m1 * v1^2) + (1/2 * m2 * v2^2)
= (1/2 * 0.0059 kg * (-0.117 m/s)^2) + (1/2 * 0.0177 kg * v2^2)
= 0.000418 J + 0.5 * 0.0177 kg * v2^2
Since the collision is elastic, the total kinetic energy before the collision is equal to the total kinetic energy after the collision. Therefore:
KE_initial = KE_final
0.00726 J = 0.000418 J + 0.5 * 0.0177 kg * v2^2
Solving for v2^2:
0.00726 J - 0.000418 J = 0.5 * 0.0177 kg * v2^2
0.006842 J = 0.00885 kg * v2^2
Dividing both sides by 0.00885 kg:
v2^2 = 0.773 J / kg
Taking the square root of both sides to solve for v2:
v2 ≈ ±0.878 m/s
Since the 5.9 g coin moves to the left, the velocity v2 must also be negative.
Therefore, v2 ≈ -0.878 m/s
Now, to find the amount of kinetic energy transferred to the 17.7 g coin, we can calculate the difference in kinetic energy before and after the collision:
KE_transferred = KE_final - KE_initial
= (0.000418 J + 0.5 * 0.0177 kg * (-0.878 m/s)^2) - 0.00726 J
= 0.00034 J
Therefore, approximately 0.00034 J of kinetic energy is transferred to the 17.7 g coin during the elastic collision.