A 5.6 g coin sliding to the right at 22.6 cm/s makes an elastic head-on collision with a 16.8 g coin that is initially at rest. After the collision, the 5.6 g coin moves to the left at 11.3 cm/s.
(a) Find the final velocity of the other coin.
cm/s
(b) Find the amount of kinetic energy transferred to the 16.8 g coin.
J
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To find the final velocity of the other coin after the collision, we can use the law of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision.
The momentum of an object is calculated by multiplying its mass with its velocity. So, we can write the equation as:
(mass1 * velocity1) + (mass2 * velocity2) = (mass1 * final velocity1) + (mass2 * final velocity2)
We know the masses and velocities of the two coins before and after the collision, except for the final velocity of the other (16.8 g) coin. Let's plug in the given values:
(5.6 g * 22.6 cm/s) + (16.8 g * 0 cm/s) = (5.6 g * -11.3 cm/s) + (16.8 g * final velocity2)
Converting the masses to kg and the velocities to m/s and solving for the final velocity2, we get:
(0.0056 kg * 0.226 m/s) + (0.0168 kg * 0 m/s) = (0.0056 kg * -0.113 m/s) + (0.0168 kg * final velocity2)
0.0012656 kg * m/s = -0.0006328 kg * m/s + (0.0168 kg * final velocity2)
Combining like terms:
0.0006328 kg * m/s = 0.0168 kg * final velocity2
final velocity2 = (0.0006328 kg * m/s) / 0.0168 kg
final velocity2 = 0.0377 m/s
Therefore, the final velocity of the other (16.8 g) coin is 0.0377 m/s to the right.
Now, let's move on to finding the amount of kinetic energy transferred to the 16.8 g coin.
The kinetic energy is given by the formula:
Kinetic Energy = (1/2) * mass * velocity^2
We can calculate the initial kinetic energy of the 16.8 g coin as:
(1/2) * 0.0168 kg * (0 m/s)^2 = 0 J (since it is initially at rest and has no kinetic energy)
The final kinetic energy of the 16.8 g coin can be calculated as:
(1/2) * 0.0168 kg * (0.0377 m/s)^2 = 1.2037 x 10^-5 J (rounding to four decimal places)
Therefore, the amount of kinetic energy transferred to the 16.8 g coin during the collision is approximately 1.2037 x 10^-5 J.