Problem 18 - College Physics

Question

\(\bullet\) A ball with a mass of 0.600 \(\mathrm{kg}\) is initially at rest. It is struck by a second ball having a mass of 0.400 \(\mathrm{kg}\) , initially moving with a velocity of 0.250 \(\mathrm{m} / \mathrm{s}\) toward the right along the \(x\) axis. After the collision, the 0.400 \(\mathrm{kg}\) ball has a velocity of 0.200 \(\mathrm{m} / \mathrm{s}\) at an angle of \(36.9^{\circ}\) above the \(x\) axis in the first quadrant. Both balls move on a frictionless, horizontal surface. (a) What are the magnitude and direction of the velocity of the 0.600 kg ball after the collision? (b) What is the change in the total kinetic energy of the two balls as a result of the collision?

Step-by-Step Solution

Answer

The 0.600 kg ball moves at 0.100 m/s, \(-53.1^{\circ}\) below x-axis. Kinetic energy decreases by 0.0015 J.

Step 1: Understand the conservation of momentum

Since the collision does not involve friction and external forces, we can apply the principle of conservation of momentum. The momentum before the collision should equal the momentum after the collision in both the x and y directions. For the x-axis: \( m_1 v_{1x} + m_2 v_{2x} = m_1 u_{1x} + m_2 u_{2x} \). For the y-axis: \( m_1 v_{1y} + m_2 v_{2y} = m_1 u_{1y} + m_2 u_{2y} \). Here, \(v\) and \(u\) are velocities, and subscripts 1 and 2 refer to the 0.600 kg and 0.400 kg balls respectively.

Step 2: Find velocities after collision for 0.400 kg ball

The velocity vector of the 0.400 kg ball is given as \(0.200\, \mathrm{m/s}\) at a \(36.9^{\circ}\) angle above the x-axis. Decompose this into components: \( u_{2x} = 0.200 \cos(36.9^{\circ})\) and \( u_{2y} = 0.200 \sin(36.9^{\circ})\). Hence, \( u_{2x} \approx 0.160 \) m/s and \( u_{2y} \approx 0.120 \) m/s.

Step 3: Apply conservation of momentum in x-direction

From the conservation of momentum in the x-direction and given initial conditions, we have: \( m_1 (0) + m_2 (0.250) = m_1 v_{1x} + m_2 (0.160) \). Rearranging gives \( 0.400 \times 0.250 = 0.600 v_{1x} + 0.400 \times 0.160 \). Solve for \( v_{1x} \): \( v_{1x} = \frac{0.400 \times 0.250 - 0.400 \times 0.160}{0.600} \approx 0.060 \) m/s.

Step 4: Apply conservation of momentum in y-direction

In the y-direction, initial momentum is zero since both balls were initially moving horizontally. \(0 = m_1 v_{1y} + m_2 (0.120) \). This simplifies to \( v_{1y} = -\frac{0.400 \times 0.120}{0.600} \approx -0.080 \) m/s.

Step 5: Calculate velocity magnitude and direction for 0.600 kg ball

Using the components found, apply Pythagorean theorem for magnitude: \( v_1 = \sqrt{v_{1x}^2 + v_{1y}^2} = \sqrt{(0.060)^2 + (-0.080)^2} \approx 0.100 \) m/s. For the direction, \( \theta = \tan^{-1}\left(\frac{v_{1y}}{v_{1x}}\right) = \tan^{-1}\left(\frac{-0.080}{0.060}\right) \approx -53.1^{\circ} \), indicating below the x-axis in the fourth quadrant.

Step 6: Calculate change in kinetic energy

Initial kinetic energy, \( KE_i = \frac{1}{2}m_2v_2^2 = \frac{1}{2} \times 0.400 \times (0.250)^2 = 0.0125 \) J. Final kinetic energy, \( KE_f = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2} \times 0.600 \times (0.100)^2 + \frac{1}{2} \times 0.400 \times (0.200)^2 = 0.003 + 0.008 \) J = \(0.011 \) J. The change in kinetic energy is \( \Delta KE = KE_f - KE_i = 0.011 - 0.0125 = -0.0015 \) J.

Key Concepts

Collision Physics

When two objects collide, they interact significantly over a very short time. In physics, a collision can be either elastic or inelastic. In an elastic collision, both kinetic energy and momentum are conserved. However, in an inelastic collision, only momentum is conserved while some kinetic energy is transformed into other forms of energy, such as heat or sound.

For our exercise, the collision occurs on a frictionless, horizontal surface, which means no external horizontal forces affect the system. Therefore, we can apply the principle of conservation of momentum. This principle states that the total momentum of a system remains constant in an isolated environment—no external forces or torques affect the system. In this context:

• The momentum before the collision equals the momentum after the collision.
• This applies to both the x and y components of momentum separately.

Understanding these basic principles helps to unravel the complex motions and interactions during and after the collision.

Kinetic Energy Change

Now that we've addressed the collision's physics, let's delve into the change in kinetic energy. Kinetic energy is the energy possessed by an object due to its motion, given by the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is velocity.

In the exercise, the initial kinetic energy is only due to the second ball (0.400 kg) since the first ball is initially at rest. We calculated the initial kinetic energy as 0.0125 J based on the second ball's velocity. After the collision, both balls move, and we computed the total final kinetic energy to be 0.011 J.

The change in kinetic energy is obtained by subtracting the initial kinetic energy from the final kinetic energy, leading to a slight loss of 0.0015 J. This loss indicates that the collision is slightly inelastic, where some energy is dissipated into other forms, even as the system's momentum is conserved.

Momentum in X and Y Directions

Momentum, a vector quantity, has both direction and magnitude. Therefore, for collisions, it's crucial to consider momentum separately in both the x and y directions. In the exercise's context:

• For the x-direction, \( 0.600 imes v_{1x} + 0.400 imes 0.160 \), must equal the initial momentum, 0.400 kg ball's momentum before the collision: \( 0.400 \times 0.250 \).
• For the y-direction, since all initial momentum is along the x-axis, \( m_1 v_{1y} + 0.400 imes 0.120 \) must equal zero.

By considering these components separately, we ensure a comprehensive understanding of how collision affects each component of momentum. The velocity components of a ball are separated into the x and y axes to analyze this conservation.

Velocity Components Calculation

Understanding how to calculate velocity components is crucial for analyzing motion in physics, especially in two-dimensional collisions. The second ball's velocity after the collision is given as 0.200 m/s at a 36.9° angle above the x-axis. This indicates that we need to find the respective x and y components of its velocity:

• The x-component (\( u_{2x} \)): \( u_{2x} = 0.200 \,\cos(36.9^{\circ}) \approx 0.160 \mathrm{m/s} \).
• The y-component (\( u_{2y} \)): \( u_{2y} = 0.200 \,\sin(36.9^{\circ}) \approx 0.120 \mathrm{m/s} \).

Further calculations give the x and y components of the other ball's final velocity. Once computed, these components are used to find the overall speed using the Pythagorean theorem, and the direction using inverse trigonometric functions.