/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 79 A ball with mass \(M,\) moving h... [FREE SOLUTION] | 91Ó°ÊÓ

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A ball with mass \(M,\) moving horizontally at 5.00 \(\mathrm{m} / \mathrm{s}\) , collides elastically with a block with mass 3 \(\mathrm{M}\) that is initially hanging at rest from the ceiling on the end of a 50.0 -m wire. Find the maximum angle through which the block swings after it is hit.

Short Answer

Expert verified
The maximum angle the block swings is approximately 6.62 degrees.

Step by step solution

01

Understand Elastic Collision

In an elastic collision, both momentum and kinetic energy are conserved. The moving ball with mass \(M\) hits a stationary block with mass \(3M\). After the collision, the ball moves with a velocity \(v'\) and the block starts moving with velocity \(V\).
02

Apply Conservation of Momentum

The initial momentum is \(M \times 5.00\ \mathrm{m/s}\). After the collision, the momentum is split between the ball and the block: \(M \times v' + 3M \times V = M \times 5.00 \). This simplifies to \(v' + 3V = 5.00\).
03

Apply Conservation of Kinetic Energy

The initial kinetic energy is \(\frac{1}{2}M \times (5.00)^2\). After collision, the energies are \(\frac{1}{2}M(v')^2 + \frac{1}{2}(3M)V^2 = \frac{1}{2}M(5.00)^2\). Simplifying, we have \((v')^2 + 3(V)^2 = 25.0\).
04

Solve the Equations

From the two equations, \(v' + 3V = 5.00\) and \((v')^2 + 3(V)^2 = 25.0\), solve for \(v'\) and \(V\). Substitute \(v' = 5.00 - 3V\) into the energy equation: \((5.00 - 3V)^2 + 3V^2 = 25.0\).Simplifying gives: \(25 - 30V + 9V^2 + 3V^2 = 25\). Thus, \(12V^2 - 30V = 0\). Factoring out \(6V\) gives \(6V(2V - 5) = 0\). Therefore, \(V = 0\) or \(V = 2.5\). Since \(V = 0\) would imply no motion, use \(V = 2.5\ \mathrm{m/s}\).
05

Relate Block's Velocity and Maximum Height

The kinetic energy at the bottom converts to potential energy at the highest point of the swing: \(\frac{1}{2}(3M)(V)^2 = (3M)gh\).Cancel \(M\) and solve for \(h\): \(0.5 \times (2.5)^2 = 9.8h\) \implies \(h = 0.3189\) meters.
06

Calculate Maximum Swing Angle

Using the height \(h\) and wire length \(L = 50.0\) m, find the maximum angle \(\theta\) using the relation:\( \cos \theta = \frac{L - h}{L} = \frac{50 - 0.3189}{50} \approx 0.9936\).The angle \(\theta\) is found by \( \theta = \arccos(0.9936)\), which yields \(\theta \approx 6.62\) degrees.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conservation of Momentum
In every elastic collision, momentum is a key player. It's conserved before and after the event occurs. This means the total momentum of the system does not change from just before collision to right after. Imagine you're holding a ball. When you throw it smoothly across a table towards another ball at rest, both balls make up a system. Although the first ball transfers some of its momentum to the second, the overall momentum of the two-ball system remains the same.

For the exercise, we have a ball and a block. The ball is moving while the block is at rest. The conservation of momentum equation here is quite simple:
  • Before the collision: Momentum = Mass of ball × Velocity of ball.
  • After the collision: Momentum = Sum of momentum of both objects.
  • Velocity relationships: Ball's velocity post-collision is involved to find the swing angle.
Essentially, by using the equation \(M \times v' + 3M \times V = M \times 5.00\), we can start to unravel the block's future motion.
Conservation of Kinetic Energy
Another important conservation law in elastic collisions is kinetic energy. Kinetic energy, like momentum, stays constant right through the crash. This means the energy that the ball has initially, because of its motion, is re-distributed between the ball and block without any loss to the environment.

What’s exciting here is how we apply this law:
  • Kinetic energy before collision: The full energy is in the moving ball.
  • After collision: Energy spreads across both the ball and block.
  • The expressions turn into: \((v')^2 + 3(V)^2 = 25.0\). This relationship helps find velocities.
It shows the beauty of physics, where even after a collision, the energy is a neatly packaged constant, guiding us to decode post-collision velocities.
Energy Conversion
Once the collision is done, how the block swings gives us a peek into the fascinating realm of energy conversion. After getting a nice kick of kinetic energy from the ball, the block starts to fly. This energy now wants to achieve something new—it wants to climb up, morphing gradually into potential energy at the swing's peak.

When the block ascends following the collision, we view two key aspects:
  • Initial kinetic energy conversion to potential energy: Simplified as \(\frac{1}{2}(3M)(V)^2 = (3M)gh\).
  • This change helps find maximum height \(h\).
  • Determine the swing’s maximum angle using trigonometric principles.
Calculating from the energy conservation, with height \(h\) found to be 0.3189 meters, connects neatly to angles using \(\cos \theta\). This connection between heights and angles gives a complete image of the motion derived entirely through energetic transformations rather than direct forces.

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