A billiard ball rolling across a table to the right at 2.2 m/s makes a head-on elastic collision with an identical ball. The mass of a billiard ball is 32 g.
If the second ball is initially at rest, what is the velocity of the first ball after the collision?
If the second ball is initially at rest, what is the velocity of the second ball after the collision?
If the second ball is initially moving to the left with a velocity of -1.1 m/s, what is the velocity of the first ball after the collision?
If the second ball is initially moving to the left with a velocity of -1.1 m/s, what is the velocity of the second ball after the collision?
1. If the second ball is initially at rest, the velocity of the first ball after the collision can be found using the formula for the final velocity of an object in an elastic collision:
v1f = ( (m1 - m2) / (m1 + m2) ) * v1i + ( (2 * m2) / (m1 + m2) ) * v2i
Since the two billiard balls have identical masses (m1 = m2) and the second ball is initially at rest (v2i = 0), the formula simplifies to:
v1f = ( (m1 - m2) / (m1 + m2) ) * v1i
v1f = ( (32 g - 32 g) / (32 g + 32 g) ) * 2.2 m/s
v1f = 0 m/s
So, the first ball will come to rest after the collision.
2. To find the velocity of the second ball after the collision, we use the same formula, but this time solving for v2f:
v2f = ( (2 * m1) / (m1 + m2) ) * v1i + ( (m2 - m1) / (m1 + m2) ) * v2i
Similarly, since m1 = m2 and v2i = 0, the formula simplifies to:
v2f = ( (2 * m1) / (m1 + m2) ) * v1i
v2f = ( (2 * 32 g) / (32 g + 32 g) ) * 2.2 m/s
v2f = 2.2 m/s
So, the second ball will have a velocity of 2.2 m/s to the right after the collision.
3. If the second ball is initially moving to the left with a velocity of -1.1 m/s, we can use the same formulas as before, but this time with v2i = -1.1 m/s:
v1f = ( (m1 - m2) / (m1 + m2) ) * v1i + ( (2 * m2) / (m1 + m2) ) * v2i
v1f = ( (32 g - 32 g) / (32 g + 32 g) ) * 2.2 m/s + ( (2 * 32 g) / (32 g + 32 g) ) * -1.1 m/s
v1f = 0 + (-1.1 m/s)
v1f = -1.1 m/s
So, the first ball will have a velocity of -1.1 m/s to the left after the collision.
4. To find the velocity of the second ball after the collision, we use the same formula as before but with v2i = -1.1 m/s:
v2f = ( (2 * m1) / (m1 + m2) ) * v1i + ( (m2 - m1) / (m1 + m2) ) * v2i
v2f = ( (2 * 32 g) / (32 g + 32 g) ) * 2.2 m/s + ( (32 g - 32 g) / (32 g + 32 g) ) * -1.1 m/s
v2f = 2.2 m/s + 0
v2f = 2.2 m/s
So, the second ball will have a velocity of 2.2 m/s to the right after the collision.
To solve problems involving elastic collisions, we can use the principle of conservation of momentum and the principle of conservation of kinetic energy.
1. If the second ball is initially at rest:
- To determine the velocity of the first ball after the collision, we notice that the initial momentum of the system is only due to the first ball. Therefore, the final momentum of the system must also come from the first ball.
- Using the principle of conservation of momentum: m1v1i = m1v1f + m2v2f (since m1 = m2)
- Since the second ball is initially at rest, its initial velocity (v2i) is 0.
- Therefore, the equation becomes: m1v1i = m1v1f
- Plugging in the values, with m1 = m2 = 32 g = 0.032 kg and v1i = 2.2 m/s, we get: 0.032 kg * 2.2 m/s = 0.032 kg * v1f
- Solving for v1f, we find that the velocity of the first ball after the collision is also 2.2 m/s.
2. If the second ball is initially at rest:
- To determine the velocity of the second ball after the collision, we use the principle of conservation of momentum and notice that the final momentum of the system is only due to the second ball.
- Using the principle of conservation of momentum: m1v1i = m1v1f + m2v2f (since m1 = m2)
- Since the second ball is initially at rest, its initial velocity (v2i) is 0.
- Therefore, the equation becomes: m1v1i = m1v1f + 0
- Plugging in the values, with m1 = m2 = 32 g = 0.032 kg and v1i = 2.2 m/s, we get: 0.032 kg * 2.2 m/s = 0.032 kg * v1f
- Solving for v1f, we find that the velocity of the first ball after the collision is 2.2 m/s.
- Since the second ball is initially at rest, its final velocity (v2f) is 0 m/s.
3. If the second ball is initially moving to the left with a velocity of -1.1 m/s:
- To determine the velocity of the first ball after the collision, we use the principle of conservation of momentum and notice that the final momentum of the system is due to both balls.
- Using the principle of conservation of momentum: m1v1i + m2v2i = m1v1f + m2v2f
- Plugging in the values, with m1 = m2 = 32 g = 0.032 kg, v1i = 2.2 m/s, and v2i = -1.1 m/s, we get: 0.032 kg * 2.2 m/s + 0.032 kg * (-1.1 m/s) = 0.032 kg * v1f + 0.032 kg * v2f
- Solving for v1f, we find that the velocity of the first ball after the collision is 0.55 m/s.
- To determine the velocity of the second ball after the collision, we can plug the known values into the equation for conservation of momentum.
- Therefore, we have: 0.032 kg * 2.2 m/s + 0.032 kg * (-1.1 m/s) = 0.032 kg * 0.55 m/s + 0.032 kg * v2f
- Solving for v2f, we find that the velocity of the second ball after the collision is -0.55 m/s.
In summary:
1. The velocity of the first ball after the collision is 2.2 m/s.
2. The velocity of the second ball after the collision is 0 m/s.
3. The velocity of the first ball after the collision is 0.55 m/s.
4. The velocity of the second ball after the collision is -0.55 m/s.
To solve these questions, we can use the principles of conservation of momentum and conservation of kinetic energy.
Conservation of momentum states that the total momentum of a system before a collision is equal to the total momentum after the collision. Mathematically, this can be written as:
m1v1i + m2v2i = m1v1f + m2v2f
where
m1 is the mass of the first ball,
m2 is the mass of the second ball,
v1i is the initial velocity of the first ball,
v2i is the initial velocity of the second ball,
v1f is the final velocity of the first ball,
v2f is the final velocity of the second ball.
Conservation of kinetic energy states that the total kinetic energy of a system before a collision is equal to the total kinetic energy after the collision. Mathematically, this can be written as:
(1/2)m1(v1i)^2 + (1/2)m2(v2i)^2 = (1/2)m1(v1f)^2 + (1/2)m2(v2f)^2
where all variables are the same as in the previous equation.
Now let's solve each question individually:
1. If the second ball is initially at rest:
Since the second ball is at rest, its initial velocity (v2i) is 0 m/s. The first ball's initial velocity (v1i) is given as 2.2 m/s. We can substitute these values into the equations above and solve for v1f and v2f:
(0.032 kg)(0 m/s) + (0.032 kg)(2.2 m/s) = (0.032 kg)(v1f) + (0.032 kg)(v2f)
(1/2)(0.032 kg)(0 m/s)^2 + (1/2)(0.032 kg)(2.2 m/s)^2 = (1/2)(0.032 kg)(v1f)^2 + (1/2)(0.032 kg)(v2f)^2
Simplifying these equations, we get:
0.0704 = 0.032v1f + 0.032v2f
0.154774 = 0.001056v1f^2 + 0.001056v2f^2
To solve for v1f and v2f, we have two equations and two unknowns. We can use algebra or a numerical method like substitution or elimination to solve these equations. The solution will give us the velocities of the balls after the collision.
2. If the second ball is initially moving to the left with a velocity of -1.1 m/s:
In this case, we still have the initial velocity of the first ball as 2.2 m/s, but now the initial velocity of the second ball (v2i) is -1.1 m/s. We can use the same equations as before:
(0.032 kg)(2.2 m/s) + (0.032 kg)(-1.1 m/s) = (0.032 kg)(v1f) + (0.032 kg)(v2f)
(1/2)(0.032 kg)(2.2 m/s)^2 + (1/2)(0.032 kg)(-1.1 m/s)^2 = (1/2)(0.032 kg)(v1f)^2 + (1/2)(0.032 kg)(v2f)^2
Simplifying and solving these equations, we will get the velocities of the balls after the collision.