Starting with an initial speed of 5.00 m/s at a height of 0.335 m, a 1.75-kg ball swings downward and strikes a 4.40-kg ball that is at rest, as the drawing shows.
(a) Using the principle of conservation of mechanical energy, find the speed of the 1.75-kg ball just before impact.
m/s
(b) Assuming that the collision is elastic, find the velocities (magnitude and direction) of both balls just after the collision.
m/s (1.75 kg-ball)
m/s (4.40 kg-ball)
(c) How high does each ball swing after the collision, ignoring air resistance?
m (1.75 kg-ball)
m (4.40 kg-ball)
(a) 4.45 m/s
(b) 2.22 m/s (1.75 kg-ball), 2.78 m/s (4.40 kg-ball)
(c) 0.67 m (1.75 kg-ball), 0.89 m (4.40 kg-ball)
To solve this problem, we will use the principles of conservation of mechanical energy and momentum.
(a) Conservation of mechanical energy tells us that the total mechanical energy before the collision is equal to the total mechanical energy after the collision. The initial mechanical energy is given by the sum of the kinetic energy and gravitational potential energy of the 1.75-kg ball:
Initial mechanical energy = 1/2 * mass * velocity^2 + mass * gravity * height
= 1/2 * 1.75 kg * (5.00 m/s)^2 + 1.75 kg * 9.8 m/s^2 * 0.335 m
The final mechanical energy after the collision is only in terms of kinetic energy because the balls are in motion:
Final mechanical energy = 1/2 * mass * velocity_after^2
Since we are assuming an elastic collision, the kinetic energy of the system is conserved. Thus, the initial mechanical energy is equal to the final mechanical energy. Therefore:
1/2 * 1.75 kg * (5.00 m/s)^2 + 1.75 kg * 9.8 m/s^2 * 0.335 m = 1/2 * 1.75 kg * velocity_after^2
Simplifying and solving for velocity_after:
velocity_after^2 = ( (5.00 m/s)^2 + 2 * 9.8 m/s^2 * 0.335 m )
velocity_after = sqrt( (5.00 m/s)^2 + 2 * 9.8 m/s^2 * 0.335 m )
Calculating this value gives us the speed of the 1.75 kg ball just before impact.
(b) For an elastic collision, momentum is conserved. Therefore, we can use the momentum equation to solve for the velocities of both balls just after the collision. Let's call the velocity of the 1.75 kg ball after the collision as v1 and the velocity of the 4.40 kg ball after the collision as v2.
Momentum before collision = Momentum after collision
(1.75 kg * 5.00 m/s) + (4.40 kg * 0 m/s) = 1.75 kg * v1 + 4.40 kg * v2
Simplifying the equation, we get:
8.75 kg m/s = 1.75 kg * v1 + 4.40 kg * v2
Since we have two variables and only one equation, we need a second equation to solve for v1 and v2. The second equation comes from the conservation of mechanical energy, which tells us that the sum of the kinetic energy of both balls after the collision is equal to the initial mechanical energy:
1/2 * 1.75 kg * v1^2 + 1/2 * 4.40 kg * v2^2 = 1/2 * 1.75 kg * (5.00 m/s)^2 + 1.75 kg * 9.8 m/s^2 * 0.335 m
Now, we have a system of two equations:
1.75 kg * v1 + 4.40 kg * v2 = 8.75 kg m/s ---- (Equation 1)
1/2 * 1.75 kg * v1^2 + 1/2 * 4.40 kg * v2^2 = 1/2 * 1.75 kg * (5.00 m/s)^2 + 1.75 kg * 9.8 m/s^2 * 0.335 m ---- (Equation 2)
We can solve this system of equations simultaneously to find the velocities v1 and v2.
(c) To determine the height each ball swings after the collision, we can use the principle of conservation of mechanical energy. The final mechanical energy after the collision is equal to the sum of the potential energy and kinetic energy of each ball:
Final mechanical energy = 1/2 * mass * v^2 + mass * gravity * height
We can use this equation to determine the height each ball swings after the collision by substituting the respective masses and velocities.
(a) To find the speed of the 1.75 kg ball just before impact using the principle of conservation of mechanical energy, we need to equate the initial potential energy to the kinetic energy just before impact. The equation for conservation of mechanical energy can be written as:
Initial potential energy = Final kinetic energy
The initial potential energy (U) of the 1.75 kg ball is given by:
U = m * g * h
where m is the mass of the ball (1.75 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the initial height (0.335 m).
Substituting the values, we get:
U = 1.75 kg * 9.8 m/s^2 * 0.335 m = 5.45075 J
Since the system is closed and there is no change in potential energy after impact, the final kinetic energy (K) is equal to the initial potential energy:
K = U = 5.45075 J
The kinetic energy (K) is given by:
K = (1/2) * m * v^2
where v is the speed of the ball just before impact.
Rearranging the equation and plugging in the values, we can solve for v:
5.45075 J = (1/2) * 1.75 kg * v^2
v^2 = (2 * 5.45075 J) / 1.75 kg
v^2 = 6.228 J/kg
v = sqrt(6.228 J/kg) ≈ 2.49 m/s
Therefore, the speed of the 1.75 kg ball just before impact is approximately 2.49 m/s.
(b) Assuming an elastic collision, the velocities (magnitude and direction) of both balls just after the collision can be found using the principle of conservation of momentum. The equation for conservation of momentum can be written as:
Initial momentum = Final momentum
The initial momentum of the system is given by:
p_initial = m1 * v1_initial + m2 * v2_initial
where m1 and m2 are the masses of the balls (1.75 kg and 4.40 kg, respectively), and v1_initial and v2_initial are the velocities of the balls just before the collision.
Since the 4.40 kg ball is at rest initially, its velocity (v2_initial) is 0 m/s.
The equation for final momentum can be written as:
p_final = m1 * v1_final + m2 * v2_final
where v1_final and v2_final are the velocities of the balls just after the collision.
Considering conservation of momentum, we have:
m1 * v1_initial = m1 * v1_final + m2 * v2_final
Substituting the values, we get:
1.75 kg * 2.49 m/s = 1.75 kg * v1_final + 4.40 kg * v2_final
Solving this equation will give us the velocities of both balls just after the collision.
(c) To find the maximum height each ball swings after the collision, we need to apply the principle of conservation of mechanical energy again. The initial kinetic energy just after the collision will be equal to the final potential energy at maximum height (assuming no energy losses due to air resistance).
The equation for conservation of mechanical energy can be written as:
Initial kinetic energy = Final potential energy
The initial kinetic energy of each ball just after the collision (K_initial) is given by:
K_initial = (1/2) * m1 * v1_final^2 + (1/2) * m2 * v2_final^2
where m1 and m2 are the masses of the balls (1.75 kg and 4.40 kg, respectively), and v1_final and v2_final are the velocities of the balls just after the collision.
Since the total mechanical energy remains constant throughout the motion, the final potential energy (U_final) at maximum height will be equal to the initial potential energy (U) before the collision.
Therefore,
U = U_final
m1 * g * h1 + m2 * g * h2 = m1 * g * H1 + m2 * g * H2
where g is the acceleration due to gravity (9.8 m/s^2), h1 and h2 are the initial heights of the balls just before the collision (0.335 m), and H1 and H2 are the maximum heights each ball swings after the collision.
Now, we have two equations and two unknowns (H1 and H2). Solving these equations will give us the maximum height each ball swings after the collision.