Two Balls of masses
mA = 45 grams and mB = 65 grams
are suspended as follows: both Balls are hanging by a 30 cm string and are vertically parallel to each other. The lighter ball is pulled away to a 66 degree angle with the vertical and is released, hitting the larger mass.
a) What is the velocity of the lighter ball before impact?
b) What is the velocity of each ball after the elastic collision?
c) What will be the maximum height of each ball after the elastic collision?
To solve this problem, we need to consider the law of conservation of momentum and the law of conservation of energy. Let's break it down step by step:
a) What is the velocity of the lighter ball before impact:
To find the velocity of the lighter ball before impact, we can use the concept of potential energy. The potential energy of an object is given by the equation: Potential Energy (PE) = mass (m) * acceleration due to gravity (g) * height (h).
In this case, the height (h) is given as 30 cm (or 0.3 meters), and the mass (m) is given as 45 grams (or 0.045 kg).
Potential Energy (PE) = 0.045 kg * 9.8 m/s^2 * 0.3 m = 0.1323 J.
Since there is no initial velocity (as it is pulled to an angle and released), all the potential energy is converted into kinetic energy before impact.
Kinetic Energy (KE) = (1/2) * mass (m) * velocity^2.
We can rearrange this equation to find the velocity:
velocity^2 = (2 * KE) / mass.
Plugging in the known values:
velocity^2 = (2 * 0.1323 J) / 0.045 kg.
velocity^2 ≈ 5.88 m^2/s^2.
Therefore, the velocity of the lighter ball before impact is approximately √5.88 m/s.
b) What is the velocity of each ball after the elastic collision:
When two objects collide elastically, the total momentum before the collision is equal to the total momentum after the collision.
Before the collision, the lighter ball has a velocity (vA) and the heavier ball has no velocity (since it is at rest).
After the collision, the lighter ball (mA) and the heavier ball (mB) will have new velocities (v1A and v1B, respectively).
The momentum of each ball is given by: momentum (p) = mass (m) * velocity (v).
So, before the collision:
p_initial = mA * vA + mB * vB = 0 (since the heavier ball is at rest).
After the collision:
p_final = mA * v1A + mB * v1B.
Since momentum is conserved, we have:
p_initial = p_final.
0 = mA * vA + mB * vB = mA * v1A + mB * v1B.
From this equation, we can solve for v1A in terms of vA and vB:
v1A = ((mA - mB) / (mA + mB)) * vA + (2 * mB / (mA + mB)) * vB.
Similarly, we can solve for v1B in terms of vA and vB:
v1B = (2 * mA / (mA + mB)) * vA - ((mA - mB) / (mA + mB)) * vB.
Plugging in the known values, we can calculate the velocities after the collision.
c) What will be the maximum height of each ball after the elastic collision:
After the elastic collision, the total mechanical energy (potential energy + kinetic energy) will be conserved.
For each ball, the maximum height can be calculated using the conservation of energy equation:
Potential Energy (PE) = mass (m) * g * height (h).
We know the mass (m) and the acceleration due to gravity (g), and we need to find the height (h) after the collision.
From the known velocities (v1A and v1B) and the conservation of momentum, we can calculate the final kinetic energy.
Then, the total mechanical energy after the collision is the sum of the final kinetic energy and the potential energy, which is equal to the maximum height.
Using these steps, we can calculate the maximum height for each ball after the collision.