A 550-g ball traveling at 8.0 m/s undergoes a head-on elastic collision with a 250-g ball traveling toward it also at 8.0 m/s. What is the speed of the 250-g mass after the collision?
You just have to use both momentum and energy equations, and do the messy algebra. Two equations, two unknowns (the two ball velocities after collision).
Enjoy the messy algebra.
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
The momentum of an object is defined as the product of its mass and velocity (p = mv).
Step 1: Calculate the momentum of the 550-g ball before the collision.
Mass of the 550-g ball (m1) = 550 g = 0.550 kg.
Velocity of the 550-g ball (v1) = 8.0 m/s.
Momentum of the 550-g ball (p1) = m1 * v1.
p1 = (0.550 kg) * (8.0 m/s)
= 4.40 kg·m/s.
Step 2: Calculate the momentum of the 250-g ball before the collision.
Mass of the 250-g ball (m2) = 250 g = 0.250 kg.
Velocity of the 250-g ball (v2) = -8.0 m/s (negative sign indicates opposite direction).
Momentum of the 250-g ball (p2) = m2 * v2.
p2 = (0.250 kg) * (-8.0 m/s)
= -2.00 kg·m/s.
Step 3: Calculate the total momentum before the collision.
Total momentum before the collision (p1_total) = p1 + p2.
p1_total = 4.40 kg·m/s + (-2.00 kg·m/s)
= 2.40 kg·m/s.
Step 4: Calculate the total mass of both balls after the collision.
Total mass after the collision (m_total) = m1 + m2.
m_total = 0.550 kg + 0.250 kg
= 0.800 kg.
Step 5: Apply the principle of conservation of momentum to find the speed of the 250-g ball after the collision.
Total momentum after the collision (p_total) = m_total * v_total.
v_total = p_total / m_total.
Given that the total momentum before the collision (p1_total) = total momentum after the collision (p_total), we can solve for the velocity of the 250-g ball (v_total).
v_total = p1_total / m_total
= 2.40 kg·m/s / 0.800 kg
= 3.00 m/s.
Therefore, the speed of the 250-g mass after the collision is 3.00 m/s.
To find the speed of the 250-g mass after the collision, we can use the principle of conservation of momentum. In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision.
Momentum (p) is calculated as the product of mass (m) and velocity (v).
Given:
Mass of the first ball (m1) = 550 g = 0.55 kg
Velocity of the first ball (v1) = 8.0 m/s
Mass of the second ball (m2) = 250 g = 0.25 kg
Velocity of the second ball (v2) = -8.0 m/s (note the negative sign indicates the opposite direction)
Using the conservation of momentum equation:
m1 * v1i + m2 * v2i = m1 * v1f + m2 * v2f
where,
v1i and v2i are the initial velocities of the first and second balls, respectively,
v1f and v2f are the final velocities of the first and second balls, respectively.
Substituting the given values, we get:
(0.55 kg * 8.0 m/s) + (0.25 kg * -8.0 m/s) = (0.55 kg * v1f) + (0.25 kg * v2f)
Simplifying the equation:
4.4 kg·m/s - 2.0 kg·m/s = 0.55 kg * v1f + 0.25 kg * v2f
2.4 kg·m/s = 0.55 kg * v1f + 0.25 kg * v2f
Now, we also know that the relative velocity of the two balls is zero after the collision (since they stick together). Hence:
v2f - v1f = 0
Rearranging the equation, we have:
v2f = v1f
Substituting this into the previous equation:
2.4 kg·m/s = 0.55 kg * v1f + 0.25 kg * v1f
2.4 kg·m/s = (0.55 kg + 0.25 kg) * v1f
2.4 kg·m/s = 0.8 kg * v1f
Now, we solve for v1f:
v1f = (2.4 kg·m/s) / (0.8 kg)
v1f = 3.0 m/s
Since v2f = v1f, the speed of the 250-g mass after the collision is also 3.0 m/s.