A 0.25-kg ball is attached to a 26-cm piece of string. The ball is first raised so that the string is taut and horizontal, then the ball is released so that, at the bottom of its swing, it undergoes an elastic head-on collision with a 0.21-kg ball that is free to roll along a horizontal table.
a) What is the speed of the swinging ball just before the collision? (Answer: 2.3 m/s)
b) What is the speed of the 0.21-kg ball just after the collision? (Answer: 2.5 m/s)
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It's an elastic collision, so I know that total energy and momentum are conserved.
I think I need to write two equations (one for kinetic energy and one for momentum), but I don't know how... :S
To solve this problem, let's start by writing the equations for conservation of energy and momentum.
a) Conservation of energy: The total energy before the collision is equal to the total energy after the collision.
The kinetic energy of the swinging ball before the collision (KE1) is given by:
KE1 = (1/2) * m1 * v1^2
where m1 is the mass of the swinging ball (0.25 kg) and v1 is its speed just before the collision (which we want to find).
The kinetic energy of the rolling ball after the collision (KE2) is given by:
KE2 = (1/2) * m2 * v2^2
where m2 is the mass of the rolling ball (0.21 kg) and v2 is its speed just after the collision.
Since it's an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision:
KE1 = KE2
Substituting the values, we get:
(1/2) * m1 * v1^2 = (1/2) * m2 * v2^2
Substituting the given values, we have:
(1/2) * 0.25 kg * v1^2 = (1/2) * 0.21 kg * v2^2
Simplifying, we get:
0.125 * v1^2 = 0.105 * v2^2
b) Conservation of momentum: The total momentum before the collision is equal to the total momentum after the collision.
The momentum of the swinging ball before the collision (p1) is given by:
p1 = m1 * v1
The momentum of the rolling ball after the collision (p2) is given by:
p2 = m2 * v2
Since it's an elastic collision, the total momentum before the collision is equal to the total momentum after the collision:
p1 = p2
Substituting the values, we get:
m1 * v1 = m2 * v2
Substituting the given values, we have:
0.25 kg * v1 = 0.21 kg * v2
Now we have two equations and two unknowns (v1 and v2). Let's solve these equations simultaneously.
Divide the two equations to eliminate the variables m1 and m2:
(0.125 * v1^2) / (0.25 * v1) = (0.105 * v2^2) / (0.21 * v2)
Simplifying, we get:
0.5 * v1 = 0.5 * v2
Canceling out the common factor, we have:
v1 = v2
So, the speed of the swinging ball just before the collision (v1) is equal to the speed of the rolling ball just after the collision (v2).
Now, we know that v2 is 2.5 m/s (as given in the question for part b).
Therefore, the speed of the swinging ball just before the collision (v1) is also 2.5 m/s, which is the answer for part a.
To solve this problem, we can use the principles of conservation of energy and momentum.
Let's start by solving part a) - finding the speed of the swinging ball just before the collision.
1. Conservation of energy:
Before the collision, the ball has potential energy due to its height, which is converted into kinetic energy at the bottom of the swing.
The potential energy at the highest point can be calculated using the formula PE = mgh, where m is the mass of the ball (0.25 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height. However, in this case, the height is zero because the string is horizontal. Therefore, the potential energy is zero.
At the bottom of the swing, all the potential energy is converted to kinetic energy, given by the formula KE = (1/2)mv^2, where v is the velocity.
Therefore, (1/2)mv^2 = KE = mgh = 0
Simplifying the equation, we get v^2 = 2gh. Since h is zero in this case, v^2 = 0, and the velocity of the swinging ball just before the collision is 0 m/s.
2. Conservation of momentum:
During the collision, the momentum of the swinging ball is transferred to the 0.21-kg rolling ball.
The momentum before the collision can be calculated using the formula p = mv, where m is the mass and v is the velocity.
The momentum after the collision is also calculated in the same way.
Since the swinging ball has a velocity of 0 m/s, its momentum before the collision is zero: p1 = m1 * 0 = 0.
The momentum after the collision can be denoted as p2.
According to the law of conservation of momentum, the initial momentum should be equal to the final momentum: p1 = p2.
Therefore, 0 = m1 * 0 + m2 * v2, where m2 is the mass of the 0.21-kg rolling ball, and v2 is its velocity just after the collision.
Now we can solve for part b) - finding the speed of the 0.21-kg ball just after the collision.
0 = 0 + m2 * v2
0 = v2
So, the velocity of the 0.21-kg ball just after the collision is 0 m/s.
In conclusion:
a) The speed of the swinging ball just before the collision is 0 m/s.
b) The speed of the 0.21-kg ball just after the collision is 0 m/s.