Question

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

Block B's speed just before hitting the floor is approximately 8.70 m/s.

Step-by-step solution

Analyze the Situation

Block A slides along a platform and collides with block B. Both blocks are involved in a horizontal motion initially, and block B falls off a height of 2.60 m. We need to find block B's velocity just before it hits the ground.

Use Conservation of Linear Momentum for a Horizontal Collision

Apply conservation of momentum in the horizontal direction for blocks A and B during the collision. Before the collision, only block A has horizontal momentum: \[ m_A v_{A_{initial}} + m_B v_{B_{initial}} = m_A v_{A_{final}} + m_B v_{B_{final}} \]Substitute the known values: \[ (2.00 \text{ kg}) \times (8.00 \text{ m/s}) = (2.00 \text{ kg}) \times (-2.00 \text{ m/s}) + (4.00 \text{ kg}) \times v_{B_{final}} \]

Solve for Block B's Final Velocity After Collision

Simplify the equation:\[ 16.00 \text{ kg m/s} = -4.00 \text{ kg m/s} + 4.00 \text{ kg} \times v_{B_{final}} \]Rearranging, \[ 20.00 \text{ kg m/s} = 4.00 \text{ kg} \times v_{B_{final}} \]Solving for \(v_{B_{final}}\), \[ v_{B_{final}} = \frac{20.00 \text{ kg m/s}}{4.00 \text{ kg}} = 5.00 \text{ m/s} \]

Determine the Time it Takes for Block B to Fall

Since block B falls from the platform vertically, we use the kinematic equation for free fall to find the time taken to hit the ground:\[ h = \frac{1}{2} g t^2 \]Given \( h = 2.60 \text{ m} \) and \( g = 9.81 \text{ m/s}^2 \), solve for \( t \):\[ t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 2.60}{9.81}} \approx 0.726 \text{ s} \]

Calculate Block B's Horizontal and Total Velocity

The horizontal speed of block B remains constant during its fall since no horizontal forces act on it. Thus, the horizontal velocity is still 5.00 m/s. Compute total velocity of block B just before it hits the floor using:\[ v_{total} = \sqrt{v_{horizontal}^2 + v_{vertical}^2} \]First, find the vertical velocity using \(v_{vertical} = g \times t\):\[ v_{vertical} = 9.81 \times 0.726 \approx 7.12 \text{ m/s} \]Then calculate the total velocity:\[ v_{total} = \sqrt{(5.00)^2 + (7.12)^2} \approx \sqrt{25 + 50.69} \approx \sqrt{75.69} \approx 8.70 \text{ m/s} \]

Key concepts

Kinematic Equations

Kinematic equations are essential tools in physics for describing the motion of objects. They let us connect various properties like velocity, time, and displacement. These equations apply when the acceleration is constant, which is a common scenario in problems involving gravity. In our exercise, block B is falling solely under the influence of gravity, making it a perfect case for 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

A horizontal collision involves two objects interacting along a single horizontal line. In such collisions, the conservation of linear momentum is especially useful to determine the outcomes.

**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 \( v_{B_{final}} \).

Substituting into the momentum equation and solving helps us find block B’s velocity immediately after the collision, which is found to be 5.00 m/s.

Velocity Calculation

Computing the velocity of an object right before it impacts the ground involves combining both its horizontal and vertical speeds. Effectively, this is about finding the resultant velocity vector just before the collision with the ground.

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.