A 5.50-kg ball, moving to the right at a velocity of +2.35 m/s on a frictionless table, collides head-on with a stationary 7.60-kg ball. Find the final velocities of the balls if the collision meet the following conditions.
(a) elastic
5.5-kg ball = Incorrect: Your answer is incorrect. m/s
7.6-kg ball = Incorrect: Your answer is incorrect. m/s
(b) completely inelastic
Incorrect: Your answer is incorrect. m/s
I will be happy to check your work. It is difficult to get a wrong answer on b), and for a) a bit of algebra is required. I suspect you are doing somehing wrong with signs.
To solve this problem, we can use the principle of conservation of momentum and kinetic energy.
(a) Elastic Collision:
In an elastic collision, both momentum and kinetic energy are conserved. The equation for conservation of momentum is given by:
m1 * v1_initial + m2 * v2_initial = m1 * v1_final + m2 * v2_final
Here, m1 and m2 are the masses of the respective balls, v1_initial and v2_initial are their initial velocities, and v1_final and v2_final are their final velocities.
Let's plug in the given values:
m1 = 5.50 kg
m2 = 7.60 kg
v1_initial = +2.35 m/s (right)
v2_initial = 0 m/s (stationary)
After the collision, the velocities of the balls will change. Let's assume v1_final and v2_final as the final velocities of the two balls.
The equation becomes:
5.50 kg * 2.35 m/s + 7.60 kg * 0 m/s = 5.50 kg * v1_final + 7.60 kg * v2_final
Simplifying further:
12.925 kg·m/s = 5.50 kg * v1_final + 7.60 kg * v2_final
Now, we have only one equation and two unknowns. To solve this, we need one more equation based on conservation of kinetic energy in an elastic collision.
In an elastic collision,
(1/2) * m1 * (v1_initial)^2 + (1/2) * m2 * (v2_initial)^2 = (1/2) * m1 * (v1_final)^2 + (1/2) * m2 * (v2_final)^2
Let's plug in the given values:
m1 = 5.50 kg
m2 = 7.60 kg
v1_initial = +2.35 m/s
v2_initial = 0 m/s
The equation becomes:
(1/2) * 5.50 kg * (2.35 m/s)^2 + (1/2) * 7.60 kg * (0 m/s)^2 = (1/2) * 5.50 kg * (v1_final)^2 + (1/2) * 7.60 kg * (v2_final)^2
Simplifying further:
(6.4375 kg·m^2/s^2) = (1.375 kg) * (v1_final)^2 + 0
Therefore, (1.375 kg) * (v1_final)^2 = 6.4375 kg·m^2/s^2
Solving for (v1_final)^2:
(v1_final)^2 = 6.4375 kg·m^2/s^2 / 1.375 kg
(v1_final)^2 = 4.68 m^2/s^2
Taking the square root of both sides:
v1_final = ±2.16 m/s
Now we have two possible values for v1_final: +2.16 m/s and -2.16 m/s. This means that after the collision, the 5.50-kg ball could either be moving to the right or to the left, depending on whether it is positive or negative.
To find the value of v2_final, we can substitute the value of v1_final (-2.16 m/s) into the first conservation of momentum equation:
5.50 kg * v1_final + 7.60 kg * v2_final = 12.925 kg·m/s
(5.50 kg)(-2.16 m/s) + (7.60 kg) * v2_final = 12.925 kg·m/s
Solving for v2_final:
-11.88 kg·m/s + (7.60 kg) * v2_final = 12.925 kg·m/s
(7.60 kg) * v2_final = 24.805 kg·m/s
v2_final = 24.805 kg·m/s / 7.60 kg
v2_final ≈ 3.26 m/s
So for an elastic collision, the final velocities of the balls are approximately:
5.50-kg ball: +2.16 m/s
7.60-kg ball: +3.26 m/s
(b) Completely Inelastic Collision:
In a completely inelastic collision, the two colliding objects stick together and move with a common final velocity.
To find the final velocity in a completely inelastic collision, we use the principle of conservation of momentum only:
m1 * v1_initial + m2 * v2_initial = (m1 + m2) * v_final
Plugging in the given values:
m1 = 5.50 kg
v1_initial = +2.35 m/s (right)
m2 = 7.60 kg
v2_initial = 0 m/s (stationary)
v_final = final velocity of the combined balls
The equation becomes:
5.50 kg * 2.35 m/s + 7.60 kg * 0 m/s = (5.50 kg + 7.60 kg) * v_final
Simplifying further:
12.925 kg·m/s = 13.10 kg * v_final
Solving for v_final:
v_final = 12.925 kg·m/s / 13.10 kg
v_final ≈ 0.9862 m/s
So for a completely inelastic collision, the final velocity of the combined balls is approximately 0.9862 m/s.