Problem 77 - University Physics with Modern Physics

Question

An 8.00 -kg block of wood sits at the edge of a frictionless table, 2.20 \(\mathrm{m}\) above the floor. A \(0.500-\mathrm{kg}\) blob of clay slides along the length of the table with a speed of 24.0 \(\mathrm{m} / \mathrm{s}\), strikes the block of wood, and sticks to it. The combined object leaves the edge of the table and travels to the floor. What horizontal distance has the combined object traveled when it reaches the floor?

Step-by-Step Solution

Answer

The combined object travels approximately 0.94 meters horizontally.

Step 1: Calculate the total mass

First, we need to calculate the total mass of the block of wood and the clay after the collision. This is simply the sum of their individual masses: \\[ m_{\text{total}} = m_{\text{block}} + m_{\text{clay}} = 8.00 \text{ kg} + 0.500 \text{ kg} = 8.50 \text{ kg} \]

Step 2: Apply conservation of momentum

Since the collision happens along a frictionless surface, the horizontal momentum is conserved. Set up the conservation of momentum equation: \\[ m_{\text{clay}} \cdot v_{\text{clay}} = m_{\text{total}} \cdot v_{\text{combo}} \] \where \( v_{\text{clay}} = 24.0 \text{ m/s} \) is the initial speed of the clay. Solving for \( v_{\text{combo}} \): \\[ (0.500 \text{ kg}) \cdot (24.0 \text{ m/s}) = (8.50 \text{ kg}) \cdot v_{\text{combo}} \] \\[ v_{\text{combo}} = \frac{12.0}{8.50} = 1.41 \text{ m/s} \]

Step 3: Calculate time of fall

Next, calculate the time it takes for the combined object to fall to the floor, which is a vertical motion problem:\\[ h = \frac{1}{2} g t^2 \] \where \( h = 2.20 \text{ m} \) is the height and \( g = 9.81 \text{ m/s}^2 \) is the acceleration due to gravity. Solving for \( t \):\\[ 2.20 = \frac{1}{2} \cdot 9.81 \cdot t^2 \] \\[ t^2 = \frac{2.20 \times 2}{9.81} \] \\[ t^2 = 0.448 \] \\[ t = \sqrt{0.448} = 0.669 \text{ s} \]

Step 4: Calculate horizontal distance traveled

The horizontal distance traveled is given by the horizontal velocity times the time of flight. Use the formula: \\[ \text{distance} = v_{\text{combo}} \cdot t \] \\[ \text{distance} = 1.41 \text{ m/s} \cdot 0.669 \text{ s} = 0.943 \text{ m} \]

Key Concepts

Projectile Motion

Projectile motion is a type of motion experienced by an object that is launched into the air with an initial velocity and then influenced only by gravity, ignoring air resistance. In this exercise, the block and clay fall under projectile motion once they travel horizontally off the table.

When analyzing projectile motion, it's crucial to separate the motion into vertical and horizontal components, as they are independent of each other.

• The horizontal component involves the speed the object travels parallel to the ground. Here, the block and clay's horizontal velocity is the result from the collision.
• The vertical component is controlled by gravity, dictating how long an object takes to fall a certain height, like the 2.20 m drop in this example.

Since the horizontal and vertical motions don't affect each other, the block and clay travel horizontally at a constant speed determined by the post-collision speed, while accelerating downward due to gravity until they reach the ground.

Frictionless Surface

A frictionless surface is a hypothetical surface where there is no resistance to sliding motion. This concept allows us to ignore the effects of friction which normally oppose motion. When considering the collision between the clay blob and the wooden block, the frictionless assumption greatly simplifies the analysis.

On a frictionless surface, conservation of momentum is straightforward because no external horizontal forces act on the system, i.e., friction does not lead to any energy loss or momentum gain. Thus, the horizontal momentum of the system before the collision (just the clay moving) equals the momentum after the collision (both clay and block moving together).

This allows us to directly relate the initial speed of the clay and the post-collision speed of the system using the equation:

• Momentum before collision: \( m_{\text{clay}} \times v_{\text{clay}} \)
• Momentum after collision: \( m_{\text{total}} \times v_{\text{combo}} \)

Without friction, the calculations efficiently account for momentum changes without additional factors complicating the physics.

Collision Physics

Collision physics involves studying interactions where objects strike each other, exchanging energies and momenta. In our exercise, the clay collides with the block and sticks to it – a classic inelastic collision.

In an inelastic collision like this:

• Kinetic energy is not conserved. Some of it transforms into other energy forms, e.g., heat or deformation energy, because objects stick together.
• Momentum is still conserved, which enables us to calculate the velocity of the combined clay and block system after their melding.

The properties of collisions dictate how we calculate post-collision speeds and are key to understanding how predictions are made in physics problems. The seamless merging of two objects denotes that mass sums while velocities are calculated from preserved momentum, forming the basis of collision physics and allowing for problem-solving even amidst energy transformations.