Two Bals. of masses
mA = 45 grams and mB = 65 grams
are suspended as shown. The lighter ball is pulled away to a 66 degree angle with the vertical and released:
-----------------------------
/| | -
/è| | |
/ | | | 30 cm
/ | | |
mA | | |
mA mB -
where è = 66 degrees, and the height that they are suspended at is 30 cm.
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?
======================================
I can break up the problem into 3 parts
First : Conservation of Energy
Second: Elastic Collision
Third : Conservation of Energy for each mass.
Two Bals. of masses
mA = 45 grams and mB = 65 grams
are suspended as shown. The lighter ball is pulled away to a 66 degree angle with the vertical and released:
=============================
---------/| | -
--------/è| | |
-------/ | | | 30 cm
------/ | | |
-----mA | | |
----------mA mB -
where è = 66 degrees, and the height that they are suspended at is 30 cm.
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?
======================================
I can break up the problem into 3 parts
First : Conservation of Energy
Second: Elastic Collision
Third : Conservation of Energy for each mass.
EDIT: FIXED FORMATTING.
Two Bals of masses
mA = 45 grams and mB = 65 grams
are suspended as shown. The lighter ball is pulled away to a 66 degree angle with the vertical and released:
==============================
'''''''''/|''|'''-''''''''''''
''''''''/è|''|'''|''''''''''''
'''''''/''|''|'''|'30 cm''''''
''''''/'''|''|'''|''''''''''''
'''''mA'''|''|'''|''''''''''''
''''''''''mA'mB''-''''''''''''
''''''''''''''''''''''''''''''
where è = 66 degrees, and the height that they are suspended at is 30 cm.
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?
======================================
I can break up the problem into 3 parts
First : Conservation of Energy
Second: Elastic Collision
Third : Conservation of Energy for each mass.
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?
First, let's address part a) of the problem:
To find the velocity of the lighter ball before impact, we can use the principle of conservation of energy. The total energy at the initial position is equal to the total energy at the final position, neglecting any dissipative forces.
The initial potential energy of the lighter ball (mA) is given by m * g * h, where m is the mass of the ball, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height at which the balls are suspended (30 cm or 0.3 m).
Therefore, the initial potential energy of the lighter ball is (0.045 kg) * (9.8 m/s^2) * (0.3 m) = 0.1323 Joules.
At the lowest point, the potential energy is converted into kinetic energy. Therefore, the final kinetic energy of the lighter ball is equal to its initial potential energy.
The kinetic energy (KE) is given by KE = 0.5 * m * v^2, where v is the velocity of the ball.
Therefore, 0.5 * (0.045 kg) * v^2 = 0.1323 J.
Simplifying the equation, we find:
v^2 = (0.1323 J) * 2 / (0.045 kg).
v^2 = 5.88 J / 0.045 kg.
v^2 = 130.67.
Taking the square root of both sides, we find:
v = √(130.67) = 11.42 m/s.
So, the velocity of the lighter ball before impact is approximately 11.42 m/s.