A 20.0 g marble sliding to the right at 58.8 cm/s overtakes and collides elastically with a 10.0 g marble moving in the same direction at 14.7 cm/s. After the collision, the 10.0 g marble moves to the right at 73.5 cm/s. Find the velocity of the 20.0 g marble after the collision.
To solve this problem, we can use the principles of conservation of momentum and conservation of kinetic energy.
1. Let's first calculate the initial momentum of the 20.0 g marble and the 10.0 g marble before the collision.
Momentum = mass × velocity
For the 20.0 g marble:
Momentum1 = (20.0 g) × (58.8 cm/s)
= 1176 g·cm/s
For the 10.0 g marble:
Momentum2 = (10.0 g) × (14.7 cm/s)
= 147 g·cm/s
2. Next, we can apply the principles of conservation of momentum and conservation of kinetic energy to find the final velocities of the marbles after the collision.
Conservation of momentum:
Momentum1 + Momentum2 = Final momentum of the system
Conservation of kinetic energy:
(1/2) × (mass1) × (velocity1)^2 + (1/2) × (mass2) × (velocity2)^2 = (1/2) × (mass1) × (v1)^2 + (1/2) × (mass2) × (v2)^2
Since the collision is elastic, the total kinetic energy before and after the collision remain the same.
3. Let's substitute the given values:
Momentum1 + Momentum2 = Final momentum of the system
1176 g·cm/s + 147 g·cm/s = (20.0 g + 10.0 g) × v2
(1/2) × (mass1) × (velocity1)^2 + (1/2) × (mass2) × (velocity2)^2 = (1/2) × (mass1) × (v1)^2 + (1/2) × (mass2) × (v2)^2
(1/2) × (20.0 g) × (58.8 cm/s)^2 + (1/2) × (10.0 g) × (14.7 cm/s)^2 = (1/2) × (20.0 g) × (v1)^2 + (1/2) × (10.0 g) × (v2)^2
4. Now we can solve the equations simultaneously to find the velocities (v1 and v2).
From the momentum equation:
1176 g·cm/s + 147 g·cm/s = (30.0 g) × v2
1323 g·cm/s = 30.0 g × v2
v2 = 1323 g·cm/s / (30.0 g)
v2 ≈ 44.1 cm/s
From the kinetic energy equation:
(1/2) × (20.0 g) × (58.8 cm/s)^2 + (1/2) × (10.0 g) × (14.7 cm/s)^2 = (1/2) × (20.0 g) × (v1)^2 + (1/2) × (10.0 g) × (44.1 cm/s)^2
Simplifying the equation gives:
(1/2) × (20.0 g) × (v1)^2 = 342514.56 g·cm^2/s^2 - 13405.88 g·cm^2/s^2
(1/2) × (20.0 g) × (v1)^2 = 329108.68 g·cm^2/s^2
10.0 g × (v1)^2 = 329108.68 g·cm^2/s^2
(v1)^2 = 329108.68 g·cm^2/s^2 / (10.0 g)
(v1)^2 ≈ 32910.87 cm^2/s^2
v1 ≈ √(32910.87 cm^2/s^2)
v1 ≈ 181.5 cm/s
Therefore, the velocity of the 20.0 g marble after the collision is approximately 181.5 cm/s.
To solve this problem, we can use the principles of conservation of momentum and conservation of kinetic energy.
1. Conservation of momentum: The total momentum before the collision is equal to the total momentum after the collision.
Let's denote the velocity of the 20.0 g marble before the collision as v1 and the velocity of the 10.0 g marble before the collision as v2. After the collision, the velocity of the 10.0 g marble is given as 73.5 cm/s.
The total momentum before the collision is the sum of the individual momenta:
m1 * v1 + m2 * v2 = m1 * v1' + m2 * v2'
Where:
m1 = mass of the 20.0 g marble = 20.0 g = 0.0200 kg
m2 = mass of the 10.0 g marble = 10.0 g = 0.0100 kg
v1 = velocity of the 20.0 g marble before the collision (unknown)
v2 = velocity of the 10.0 g marble before the collision = 14.7 cm/s
v1' = velocity of the 20.0 g marble after the collision (unknown)
v2' = velocity of the 10.0 g marble after the collision = 73.5 cm/s
2. Conservation of kinetic energy: The total kinetic energy before the collision is equal to the total kinetic energy after the collision.
The kinetic energy of an object is given by the formula:
KE = (1/2) * m * v^2
The total kinetic energy before the collision is the sum of the individual kinetic energies:
(1/2) * m1 * v1^2 + (1/2) * m2 * v2^2 = (1/2) * m1 * v1'^2 + (1/2) * m2 * v2'^2
Now we have two equations based on the principles of conservation of momentum and conservation of kinetic energy. We can solve them simultaneously to find the values of v1 and v1'.
Let's plug in the given values and solve for v1 and v1':
(0.0200 kg)(v1) + (0.0100 kg)(14.7 cm/s) = (0.0200 kg)(v1') + (0.0100 kg)(73.5 cm/s) -- (Equation 1)
(1/2)(0.0200 kg)(v1^2) + (1/2)(0.0100 kg)(14.7 cm/s)^2 = (1/2)(0.0200 kg)(v1'^2) + (1/2)(0.0100 kg)(73.5 cm/s)^2 -- (Equation 2)
Now we have a system of two equations with two unknowns. We can solve these equations using algebraic methods or numerical methods to find the velocities v1 and v1'.
By substituting the given values and solving the equations, we can find the solution.
Note: To convert cm/s to m/s, divide the value by 100.