Chapter 8: Problem 74
Block \(B\) (mass 4.00 kg) is at rest at the edge of a smooth platform, 2.60 m above the floor. Block \(A\) (mass 2.00 kg) is sliding with a speed of 8.00 m/s along the platform toward block \(B\). \(A\) strikes \(B\) and rebounds with a speed of 2.00 m/s. The collision projects \(B\) horizontally off the platform. What is the speed of \(B\) just before it strikes the floor?
Short Answer
Step by step solution
Analyze the Situation
Use Conservation of Linear Momentum for a Horizontal Collision
Solve for Block B's Final Velocity After Collision
Determine the Time it Takes for Block B to Fall
Calculate Block B's Horizontal and Total Velocity
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Kinematic Equations
For block B, which falls from a 2.60 m high platform, we use the equation: \[ h = \frac{1}{2} g t^2 \] This equation helps us determine the time it takes for the block to fall to the ground. Plug in the values for the height (\( h = 2.60 \text{ m} \)) and the acceleration due to gravity (\( g = 9.81 \text{ m/s}^2 \)) to find the time \( t \).
Solving yields: \[ t = \sqrt{\frac{2 \times 2.60}{9.81}} \approx 0.726 \text{ s} \] This calculated time is crucial for finding other aspects of block B’s motion, such as its final velocity when it strikes the ground.
Horizontal Collision
**Conservation of Momentum** is a fundamental principle: The total momentum before the collision equals the total momentum after, assuming no external forces act on the objects. For block A and block B, the scenario before the collision is: \[ m_A v_{A_{initial}} + m_B v_{B_{initial}} = m_A v_{A_{final}} + m_B v_{B_{final}} \] In our case, block A strikes block B which is initially at rest.
- Before the collision: Block A's momentum is \( 2.00 \text{ kg} \times 8.00 \text{ m/s} \)
- After the collision: Block A's momentum changes to \( 2.00 \text{ kg} \times (-2.00 \text{ m/s}) \) and block B gains velocity \( v_{B_{final}} \).
Velocity Calculation
Block B, post-collision, already had a horizontal velocity of 5.00 m/s. As block B descends, it accelerates vertically due to gravity. We calculate the vertical velocity \( v_{vertical} \) using: \[ v_{vertical} = g \times t \] where \( g = 9.81 \text{ m/s}^2 \) and \( t \approx 0.726 \text{ s} \). This results in \( v_{vertical} \approx 7.12 \text{ m/s} \).
To find the total velocity \( v_{total} \), we apply the Pythagorean theorem to find the magnitude of the velocity vector: \[ v_{total} = \sqrt{v_{horizontal}^2 + v_{vertical}^2} \] Substituting the values: \[ v_{total} = \sqrt{(5.00)^2 + (7.12)^2} \approx 8.70 \text{ m/s} \] Thus, block B's velocity right before hitting the ground is approximately 8.70 m/s.