A 6.0 g coin moving to the right at 21.0 cm/s makes an elastic head-on collision with a 20.0 g coin that is initially at rest. After the collision, the 6.0 g coin moves to the left at 12.5 cm/s.
a.) Find the final velocity of the other coin in centimeters.
b.) Find the amount of kinetic energy transferred to the 20.0 g coin.
To solve this problem, we can use the principles of conservation of momentum and kinetic energy.
a) Let's assume the final velocity of the 20.0 g coin is v2 (to the right).
Using the conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.
The initial momentum is given by:
(6.0 g) * (21.0 cm/s) = (6.0 g) * (-12.5 cm/s) + (20.0 g) * v2
Simplifying the equation, we get:
6.0 g * 21.0 cm/s = -6.0 g * 12.5 cm/s + 20.0 g * v2
Using consistent units (grams cancel out cm/s), we can simplify further:
6.0 * 21.0 = -6.0 * 12.5 + 20.0 * v2
126 = -75 + 20v2
20v2 = 201
v2 = 10.05 cm/s
Therefore, the final velocity of the other coin (20.0 g) is 10.05 cm/s to the right.
b) To find the amount of kinetic energy transferred to the 20.0 g coin, we need to calculate the initial kinetic energy of the 20.0 g coin and the final kinetic energy of the 6.0 g coin.
The initial kinetic energy of the 20.0 g coin is given by:
(1/2) * (20.0 g) * 0^2 = 0 J (since it is initially at rest)
The final kinetic energy of the 6.0 g coin is given by:
(1/2) * (6.0 g) * (-12.5 cm/s)^2 = 4.688 J
The amount of kinetic energy transferred to the 20.0 g coin is the difference between the final and initial kinetic energy:
Kinetic energy transferred = Final kinetic energy of the 6.0 g coin - Initial kinetic energy of the 20.0 g coin
Kinetic energy transferred = 4.688 J - 0 J
Kinetic energy transferred = 4.688 J
Therefore, the amount of kinetic energy transferred to the 20.0 g coin is 4.688 J.
To find the final velocity of the other coin, we can use the law of conservation of momentum. According to this law, the total momentum before the collision is equal to the total momentum after the collision.
The momentum of an object is given by the product of its mass and velocity. Therefore, we can write the equation for the conservation of momentum as:
(mass 1 * velocity 1) + (mass 2 * velocity 2) = (mass 1 * final velocity 1) + (mass 2 * final velocity 2)
In this case, the 6.0 g coin is moving to the right at 21.0 cm/s before the collision and to the left at 12.5 cm/s after the collision. The 20.0 g coin is initially at rest, so its velocity before the collision is 0 cm/s. Let's plug the values into the equation and solve for the final velocity of the other coin (final velocity 2).
(6.0 g * 21.0 cm/s) + (20.0 g * 0 cm/s) = (6.0 g * (-12.5 cm/s)) + (20.0 g * final velocity 2)
Now, let's convert the units to kilograms to make the calculation easier. 1 g = 0.001 kg, so we have:
(0.006 kg * 21.0 cm/s) + (0.02 kg * 0 cm/s) = (0.006 kg * (-12.5 cm/s)) + (0.02 kg * final velocity 2)
0.126 kg*cm/s = -0.075 kg*cm/s + 0.02 kg * final velocity 2
0.126 kg*cm/s + 0.075 kg*cm/s = 0.02 kg * final velocity 2
0.201 kg*cm/s = 0.02 kg * final velocity 2
final velocity 2 = 0.201 kg*cm/s / 0.02 kg
final velocity 2 = 10.05 cm/s
So, the final velocity of the 20.0 g coin is 10.05 cm/s.
Now, let's move on to finding the amount of kinetic energy transferred to the 20.0 g coin.
The kinetic energy of an object is given by the equation:
Kinetic energy = (1/2) * mass * velocity^2
Before the collision, the 20.0 g coin is at rest, so its kinetic energy is 0. After the collision, the 6.0 g coin is moving to the left at 12.5 cm/s, so its kinetic energy is:
Kinetic energy = (1/2) * 6.0 g * (12.5 cm/s)^2
Let's calculate this value:
Kinetic energy = (1/2) * 0.006 kg * (12.5 cm/s)^2
Kinetic energy = 0.000375 J
Therefore, 0.000375 J of kinetic energy is transferred to the 20.0 g coin during the collision.
m1= 6•10^-3 kg, v1 = 0.21 m/s,
m2 = 20•10^-3 kg, v2 = 0,
u1 =0.125 m/s,
u1 =? KE2 =?
m1•v1 = - m1•u1 +m2•u2,
u2 =2•m1•v1/(m1+m2) = 2•6• 10^-3•0.21/26•10^-3 =0.097 m/s
KE2 =m2•u2²/2 =20•10^-3•(0.097)²/2 =9.41•10^-5 J.