Question

A ball with mass \(M\), moving horizontally at 4.00 m/s, collides elastically with a block with mass 3\(M\) that is initially hanging at rest from the ceiling on the end of a 50.0-cm wire. Find the maximum angle through which the block swings after it is hit.

Short answer

The maximum angle \(\theta\) through which the block swings is approximately 29.3 degrees.

Step-by-step solution

Understand Elastic Collision

In an elastic collision, both momentum and kinetic energy are conserved. The ball with mass \(M\) is moving at 4.00 m/s and collides with a block mass \(3M\) initially at rest. We'll use conservation laws to find the velocity of the block after the collision.

Apply Conservation of Momentum

The total momentum before collision equals total momentum after. Let \(v_1\) be the final velocity of the ball and \(v_2\) be the final velocity of the block. The equation is: \[ M imes 4.00 = M imes v_1 + 3M imes v_2. \] This simplifies to: \[ 4 = v_1 + 3v_2. \]

Apply Conservation of Kinetic Energy

The total kinetic energy is also conserved. Thus: \[ \frac{1}{2}M \times (4.00)^2 = \frac{1}{2}M \times v_1^2 + \frac{1}{2} \times 3M \times v_2^2. \] This simplifies to: \[ 16 = v_1^2 + 3v_2^2. \]

Solve Simultaneous Equations

Solve equations from Step 2 and Step 3:From Step 2: \[ v_1 = 4 - 3v_2. \]Substitute \(v_1\) in Step 3 equation: \[ 16 = (4 - 3v_2)^2 + 3v_2^2. \]Solve this quadratic equation for \(v_2\).

Calculate Block's Velocity After Collision

Solving the quadratic equation from Step 4 gives \(v_2 = 2 \). The block's velocity after the collision is 2 m/s.

Determine Maximum Height Using Energy Conservation

The kinetic energy of the block converts into potential energy at its maximum height. Using energy conservation: \[\frac{1}{2} \times 3M \times 2^2 = 3Mgh.\]Solving for \(h\) gives: \[h = \frac{4}{2g} = \frac{2}{9.8}.\]

Calculate Maximum Angle of Swing

The maximum height \(h\) is reached when the block swings to its maximum angle \(\theta\). Using trigonometry: \[ L - h = L\cos(\theta) \] where \(L = 0.5\) m,\[0.5 - \frac{2}{9.8} = 0.5\cos(\theta).\]Solving this gives the value of \(\theta\):\[\theta = \cos^{-1}\left(\frac{0.5 - 0.204}{0.5}\right).\]

Key concepts

Conservation of Momentum

Understanding conservation of momentum is crucial when dealing with elastic collisions. Momentum is defined as the product of an object's mass and its velocity. In any isolated system, the total momentum before a collision must be equal to the total momentum after the collision. This principle helps us predict the results of interactions in closed systems, like our ball and block scenario.

Here's how it applies to elastic collisions:

• Before the collision, the ball has a mass ( \(M\) ) and a velocity of \(4.00 \, \text{m/s}\). The block is at rest, so its initial velocity is 0.
• After the collision, both objects might have different velocities. We assign \(v_1\) to the ball and \(v_2\) to the block.
• The conservation equation can be set up as \(M \times 4.00 = M \times v_1 + 3M \times v_2\), which simplifies to \(4 = v_1 + 3v_2\).

By solving this equation, we can find a relationship between the final velocities of the ball and block. Momentum conservation ensures an even exchange of energy, allowing us to track the movement in multi-object systems effectively.

Conservation of Kinetic Energy

In elastic collisions, not only is momentum conserved, but kinetic energy is also conserved. Kinetic energy is the energy that an object possesses due to its motion, calculated as \(\frac{1}{2} \times \text{mass} \times \text{velocity}^2\). This concept implies that the total kinetic energy before the collision equals the total kinetic energy after.

Let's see how it applies here:

• Before the collision, the ball has kinetic energy calculated by \(\frac{1}{2}M \times (4.00)^2 = 8M\, \text{J}\).
• After the collision, the energies are now distributed between the ball and the block. We set up the equation \(8M = \frac{1}{2}M \times v_1^2 + \frac{1}{2} \times 3M \times v_2^2\), simplifying to \(16 = v_1^2 + 3v_2^2\).

By solving this equation alongside the momentum equation, we find each object's final velocity. This dual conservation leads to precise predictions about the movement post-collision, ensuring no energy is lost in the process.

Trigonometry in Physics

Trigonometry often finds its use in physics to solve problems involving angles and lengths, particularly in circuits, waves, and motion-related problems like swinging objects. When the block is set in motion after collision, it swings in an arc, reaching a maximum height where its potential energy is highest. This link between motion and height involves trigonometric elements.

For our exercise:

• The swing's maximum height can be calculated using energy conservation, where kinetic energy is transformed into potential energy at the swing's peak: \(\frac{1}{2} \times 3M \times 2^2 = 3Mgh\), leading to \(h = \frac{2}{9.8}\)\, \text{m}.
• To find the swing's angle ( \(\theta\)), we use the length of the string ( \(L = 0.5\)\, \text{m}) and the height \(h\), following the equation \(0.5 - h = 0.5 \cos(\theta)\).
• Solving \(\cos(\theta) = \frac{0.296}{0.5}\), we find \(\theta = \cos^{-1}\left(0.592\right)\).

This trigonometric relationship allows us to translate the block's vertical and angular displacement, showcasing the practical use of trigonometry in describing motion.