/*! 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 60 block \(A\) (mass \(1.6 \mathrm{... [FREE SOLUTION] | 91Ó°ÊÓ

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block \(A\) (mass \(1.6 \mathrm{~kg}\) ) slides into block \(B\) (mass \(2.4 \mathrm{~kg}\) ), along a frictionless surface. The directions of three velocities before \((i)\) and after \((f)\) the collision are indicated; the corresponding speeds are \(v_{A i}=5.5 \mathrm{~m} / \mathrm{s}\) \(v_{B i}=2.5 \mathrm{~m} / \mathrm{s},\) and \(v_{B f}=4.9 \mathrm{~m} / \mathrm{s} .\) What are the (a) speed and (b) direction (left or right) of velocity \(\vec{v}_{A f} ?\) (c) Is the collision elastic?

Short Answer

Expert verified
The speed of Block A is 1.9 m/s to the right, and the collision is elastic.

Step by step solution

01

Initial Directions

Since direction is required, assume right as positive direction and left as negative. It is given that Velocity of Block A initially (\(v_{A i}\) ) is heading towards Block B (right direction) and Velocity of Block B after collision (\(v_{B f}\) ) is positive.
02

Apply Conservation of Momentum

The law of conservation of momentum in a closed system states that the total momentum before the collision is equal to the total momentum after the collision. The formula is: \[ m_A v_{A i} + m_B v_{B i} = m_A v_{A f} + m_B v_{B f} \]Now substitute the known values: \[1.6 \times 5.5 + 2.4 \times 2.5 = 1.6 \times v_{A f} + 2.4 \times 4.9 \]Solve for \(v_{A f}\).
03

Solve for Final Velocity of Block A

Calculate the momentum values substituting into formula from step 2: \[8.8 + 6.0 = 1.6v_{A f} + 11.76\] Combining the terms results in \[14.8 = 1.6v_{A f} + 11.76\] Subtract 11.76 from both sides: \[3.04 = 1.6v_{A f}\] Divide by 1.6 to solve for \(v_{A f}\) : \[v_{A f} = \frac{3.04}{1.6}\] Thus, \[v_{A f} = 1.9 \text{ m/s}\].
04

Determine the Direction

Since the solution for \(v_{A f}\) is positive, Block A is moving to the right after the collision.
05

Evaluate Elasticity of the Collision

An elastic collision conserves kinetic energy. Check the total initial and final kinetic energies:The kinetic energy formula is: \[ KE = \frac{1}{2} m v^2 \]Calculate initial kinetic energy:\[ KE_i = \frac{1}{2}(1.6)(5.5)^2 + \frac{1}{2}(2.4)(2.5)^2 = 24.2 + 7.5= 31.7 \text{ J} \]Calculate final kinetic energy:\[ KE_f = \frac{1}{2}(1.6)(1.9)^2 + \frac{1}{2}(2.4)(4.9)^2 = 2.888 + 28.812 = 31.7 \text{ J} \]Since \(KE_i = KE_f\), the collision is elastic.

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Key Concepts

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

Collision Elasticity
In the world of physics, collisions are often categorized based on their elasticity. Elastic collisions are a type where no kinetic energy is lost. This means that the total kinetic energy of the system (of colliding objects) before and after the collision remains the same. It's critical to understand this when analyzing interactions between objects.

To determine if a collision is elastic:
  • Compute the initial kinetic energy before the collision.
  • Calculate the final kinetic energy after the collision.
  • If both values are equal, the collision is elastic.

In this exercise, by using the kinetic energy formula, we calculated that the kinetic energy before and after the collision for blocks A and B were both 31.7 J. This equality confirms the collision is indeed elastic. This tells us that not only was momentum conserved, but so was energy.
Kinetic Energy Calculation
Kinetic energy is one of the most fundamental concepts in physics. It's the energy that an object possesses due to its motion. The formula to calculate kinetic energy is:
\[ KE = \frac{1}{2} m v^2 \]
where \(m\) is the mass of the object, and \(v\) is its velocity.

When calculating kinetic energy, it's important to:
  • Carefully substitute the mass and velocity into the formula.
  • Make sure all units are consistent, typically in kilograms for mass and meters per second for velocity.

In the case of the blocks A and B, we determined each block's initial and final kinetic energies. The initial values were \(24.2 \text{ J}\) for block A and \(7.5 \text{ J}\) for block B, summing up to \(31.7 \text{ J}\). After the collision, we find the same total: \(2.888 \text{ J}\) for A and \(28.812 \text{ J}\) for B. These computations reaffirm the collision was elastic.
Frictionless Surface Dynamics
Analyzing motion on a frictionless surface allows us to focus on core concepts like momentum without the complication of opposing forces, such as friction. Without friction, once an object is in motion, it will continue moving at a constant velocity unless acted on by an external force. This provides a simplified scenario to study collision dynamics.

On a frictionless surface:
  • Conservation of momentum can be directly applied.
  • The role of external forces is minimal, simplifying calculations.
  • Collisions become easier to analyze, particularly elastic collisions, where kinetic energy conservation can be assessed without energy loss to friction.

This exercise benefits from such conditions because it allows us to apply the principles of conservation of momentum convincingly. With no friction, shifts in velocities of the blocks A and B are purely due to their interaction, confirming how their energies and momenta change, or in this case, remain constant, elucidating the nature of their collision.

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Most popular questions from this chapter

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