A ball collides with a second ball of equal mass (2.3 kg) that is initially at rest. If the first ball is initially moving with a velocity of 3 m/s and is no longer moving after the collision, (1) what is the velocity of the second ball after the collision? (2) What is the second ball’s kinetic energy after the collision?
The correct answer and explanation is:
Let’s analyze the problem step by step.
Given:
- Mass of ball 1, m1=2.3 kgm_1 = 2.3 \, \text{kg}
- Mass of ball 2, m2=2.3 kgm_2 = 2.3 \, \text{kg} (equal mass)
- Initial velocity of ball 1, v1i=3 m/sv_{1i} = 3 \, \text{m/s}
- Initial velocity of ball 2, v2i=0 m/sv_{2i} = 0 \, \text{m/s} (at rest)
- Final velocity of ball 1, v1f=0 m/sv_{1f} = 0 \, \text{m/s} (no longer moving)
(1) What is the velocity of the second ball after the collision?
This is a classic elastic collision problem where two balls of equal mass collide, and one is initially at rest.
Key principle: Conservation of momentum m1v1i+m2v2i=m1v1f+m2v2fm_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}
Plug in values: 2.3×3+2.3×0=2.3×0+2.3×v2f2.3 \times 3 + 2.3 \times 0 = 2.3 \times 0 + 2.3 \times v_{2f}
Simplify: 6.9=2.3×v2f6.9 = 2.3 \times v_{2f}
Solve for v2fv_{2f}: v2f=6.92.3=3 m/sv_{2f} = \frac{6.9}{2.3} = 3 \, \text{m/s}
(2) What is the second ball’s kinetic energy after the collision?
Kinetic energy (KE) is given by: KE=12mv2KE = \frac{1}{2} m v^2
Calculate: KE=12×2.3×(3)2=1.15×9=10.35 JoulesKE = \frac{1}{2} \times 2.3 \times (3)^2 = 1.15 \times 9 = 10.35 \, \text{Joules}
Explanation (300 words):
This problem demonstrates the conservation of momentum and kinetic energy in collisions, specifically an elastic collision between two equal masses. The first ball, moving at 3 m/s, strikes the second ball, which is initially at rest. After the collision, the first ball stops moving, and the second ball takes on the motion.
The principle of conservation of momentum states that in a closed system with no external forces, the total momentum before and after the collision remains constant. Here, the total momentum initially is the momentum of the first ball because the second ball is at rest. After the collision, the first ball’s velocity is zero, so all the momentum must be transferred to the second ball.
Mathematically, this means: m1v1i=m2v2fm_1 v_{1i} = m_2 v_{2f}
Given equal masses, the second ball’s final velocity equals the initial velocity of the first ball, which is 3 m/s. This aligns with everyday observations of billiard balls colliding, where one ball stops, and the other moves off with the original speed.
Next, kinetic energy is the energy associated with motion. Since the first ball comes to rest, its kinetic energy is zero post-collision. The second ball gains kinetic energy equal to the initial kinetic energy of the first ball, assuming an ideal elastic collision with no energy lost to sound, heat, or deformation.
Calculating the kinetic energy of the second ball confirms this: KE=12mv2=12×2.3×9=10.35 JKE = \frac{1}{2} m v^2 = \frac{1}{2} \times 2.3 \times 9 = 10.35 \, \text{J}
This result shows that energy and momentum transfer perfectly from the first ball to the second.
In real-world collisions, some energy might be lost, but this ideal case is an important foundation in classical mechanics, illustrating fundamental conservation laws in physics.