A 49.0 g marble moving at 2.20 m/s strikes a 22.0 g marble at rest. What is the speed of each marble immediately after the collision?
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
The momentum of an object is equal to the product of its mass and velocity. Mathematically, momentum (p) is given by:
p = m * v
Where:
p is the momentum,
m is the mass of the object, and
v is the velocity of the object.
Before the collision, the first marble has a mass of 49.0 g and a velocity of 2.20 m/s, while the second marble is at rest with a mass of 22.0 g.
The total momentum before the collision is given by:
p_total_initial = (m1 * v1) + (m2 * v2)
Let's calculate the total momentum before the collision:
m1 = 49.0 g = 0.049 kg (converting grams to kilograms)
v1 = 2.20 m/s
m2 = 22.0 g = 0.022 kg
v2 = 0 m/s (at rest)
p_total_initial = (0.049 kg * 2.20 m/s) + (0.022 kg * 0 m/s)
p_total_initial = 0.1078 kg m/s
Now, let's calculate the velocity of the marbles after the collision.
As per the conservation of momentum principle, the total momentum after the collision is also equal to 0.1078 kg m/s.
Let's assume the velocity of the first marble after the collision is v1' and the velocity of the second marble is v2'.
The total momentum after the collision is given by:
p_total_final = (m1 * v1') + (m2 * v2')
Since the second marble was initially at rest (v2 = 0 m/s), the equation becomes:
p_total_final = (m1 * v1') + (m2 * 0)
Since the total momentum before and after the collision is the same, we have:
0.1078 kg m/s = (0.049 kg * v1') + (0.022 kg * 0)
Simplifying the equation:
0.1078 kg m/s = 0.049 kg * v1'
v1' = 0.1078 kg m/s / 0.049 kg
v1' = 2.2 m/s
Therefore, the velocity of the first marble after the collision (v1') is 2.2 m/s.
Since the second marble was initially at rest, its velocity after the collision (v2') is equal to the total momentum after the collision divided by its mass:
v2' = p_total_final / m2
v2' = 0.1078 kg m/s / 0.022 kg
v2' ≈ 4.90 m/s
Therefore, the velocity of the second marble after the collision (v2') is approximately 4.90 m/s.
To determine the speed of each marble immediately after the collision, we can apply the principles of conservation of momentum and kinetic energy.
The principle of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision, as long as no external forces are acting on the system. Mathematically, this can be expressed as:
(m1 * v1) + (m2 * v2) = (m1 * v1') + (m2 * v2')
Where:
m1 and m2 are the masses of the marbles
v1 and v2 are the velocities of the marbles before the collision
v1' and v2' are the velocities of the marbles after the collision
Similarly, 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. Mathematically, this can be expressed as:
(1/2 * m1 * v1^2) + (1/2 * m2 * v2^2) = (1/2 * m1 * v1'^2) + (1/2 * m2 * v2'^2)
Now let's plug in the given values:
m1 = 49.0 g = 0.049 kg (mass of the first marble)
v1 = 2.20 m/s (velocity of the first marble)
m2 = 22.0 g = 0.022 kg (mass of the second marble)
v2 = 0 m/s (velocity of the second marble at rest)
To solve the equations, we need to solve both equations simultaneously.
Let's first solve for v1':
From the conservation of momentum equation:
(m1 * v1) + (m2 * v2) = (m1 * v1') + (m2 * v2')
(0.049 kg * 2.20 m/s) + (0.022 kg * 0 m/s) = (0.049 kg * v1') + (0.022 kg * v2')
0.1078 kg m/s = 0.049 kg * v1'
v1' = 0.1078 kg m/s / 0.049 kg
v1' ≈ 2.20 m/s
Now let's solve for v2':
From the conservation of kinetic energy equation:
(1/2 * m1 * v1^2) + (1/2 * m2 * v2^2) = (1/2 * m1 * v1'^2) + (1/2 * m2 * v2'^2)
(1/2 * 0.049 kg * (2.20 m/s)^2) + (1/2 * 0.022 kg * (0 m/s)^2) = (1/2 * 0.049 kg * (2.20 m/s)^2) + (1/2 * 0.022 kg * v2'^2)
0.2702 J = 0.2702 J + (1/2 * 0.022 kg * v2'^2)
0 = (1/2 * 0.022 kg * v2'^2)
0 = v2'^2
Since there is no solution for v2'^2, it means that the second marble remained at rest after the collision.
Therefore, the speed of the first marble (m1) immediately after the collision is still 2.20 m/s, and the speed of the second marble (m2) immediately after the collision is 0 m/s.