/*! 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 57 A bullet of mass \(0.01 \mathrm{... [FREE SOLUTION] | 91Ó°ÊÓ

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A bullet of mass \(0.01 \mathrm{~kg}\) and travelling at a speed of \(500 \mathrm{~ms}^{-1}\) strikes a block of mass \(2 \mathrm{~kg}\) which is suspended by a string of length \(5 \mathrm{~m}\). The centre of gravity of the block is found to raise a vertical distance of \(0.2 \mathrm{~m}\). What is the speed of the bullet after it emerges from the block? (1) \(15 \mathrm{~ms}^{-1}\) (2) \(20 \mathrm{~ms}^{-1}\) (3) \(100 \mathrm{~ms}^{-1}\) (4) \(50 \mathrm{~ms}^{-1}\)

Short Answer

Expert verified
The speed of the bullet after passing through the block is approximately 100 m/s.

Step by step solution

01

Understand the Problem

We need to find the speed of the bullet after it strikes and passes through the block. The bullet has an initial mass and speed, and the block is suspended, allowing it to swing. We know the center of gravity of the block rises by 0.2 m. Given these details, we will use the conservation of momentum and energy principles.
02

Apply Conservation of Momentum

Since no external horizontal forces are acting, the horizontal momentum before the collision is equal to the horizontal momentum after the collision. Let \(v_b\) be the speed of the bullet after passing through the block. The equation for conservation of momentum is: \[ m_b u_b + m_B u_B = m_b v_b + m_B v_B \] Where: \(m_b = 0.01 \, \text{kg}\) (mass of bullet), \(u_b = 500 \, \text{m/s}\) (initial speed of bullet), \(m_B = 2 \, \text{kg}\) (mass of block), \(u_B = 0 \, \text{m/s}\) (initial speed of block), \(v_B\) is the speed of the block after the collision.
03

Use Conservation of Energy for the Block

The potential energy gained by the block as its center of gravity rises comes from the kinetic energy of the block after being shot. Calculate the speed of the block (\(v_B\)) using energy principles: The potential energy gain is \(m_B g h\) where \(h = 0.2 \, \text{m}\) and \(g = 9.8 \, \text{m/s}^2\).The kinetic energy of the block just after the bullet passes is \(\frac{1}{2} m_B v_B^2\).Equating the change in potential to kinetic energy: \[ m_B g h = \frac{1}{2} m_B v_B^2 \] Solving for \(v_B\): \[ v_B = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 0.2} \approx 1.98 \, \text{m/s} \]
04

Calculate the Speed of the Bullet After Collision

Substitute the value of \(v_B\) into the momentum equation from Step 2 to solve for \(v_b\): \[ 0.01 \, \text{kg} \times 500 \, \text{m/s} + 0 = 0.01 \, \text{kg} \times v_b + 2 \, \text{kg} \times 1.98 \, \text{m/s} \] Rearrange to solve for \(v_b\): \[ 5 = 0.01 v_b + 3.96 \] \[ 0.01 v_b = 1.04 \] \[ v_b = 104 \, \text{m/s} \] Thus, the speed of the bullet after it emerges from the block is approximately \(100 \, \text{m/s}\), which corresponds to option (3).

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Key Concepts

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

Conservation of Energy
The principle of conservation of energy states that energy cannot be created or destroyed, only transferred or transformed from one form to another. In this problem, the concept is highlighted through the transformation of the bullet's kinetic energy into the potential energy of the block.

When the bullet strikes the block, it causes the block to swing upward. This swing means that kinetic energy (from the bullet) is converted into potential energy (as the block rises to a higher height).
  • Initially, the block is at rest, meaning its potential energy at the initial height is zero.
  • The bullet imparts some of its kinetic energy to the block, causing it to move.
  • As the block reaches its highest point, all of that imparted kinetic energy is now stored as potential energy.
The energy is transferred using the formula for potential energy: \[ m_B g h = \frac{1}{2} m_B v_B^2 \]
Here, \(m_B\) is the block's mass, \(g\) is the acceleration due to gravity, and \(h\) is the height it rises (0.2 m in this case). By conveying how energy changes form, students can better understand the energy interactions taking place in this scenario.
Collision Dynamics
Collision dynamics explores how objects behave when they collide, focusing on momentum and energy transfers. In this exercise, a bullet collides with a suspended block, leading us to analyze the dynamics of this collision.

We assess this by invoking conservation of momentum, a fundamental principle of physics stating that the total momentum of a closed system remains constant if no external forces act on it. Here’s how it relates to our problem:
  • The initial momentum of the bullet is calculated with its mass and velocity: \(m_b u_b\).
  • The block initially has no momentum since it is at rest: \(m_B u_B = 0\).
  • After impact, the momentum is shared between the bullet and the block.
  • We use the equation: \[ m_b u_b + m_B u_B = m_b v_b + m_B v_B \]which balances the momentum before and after collision.
This equation allows us to determine the bullet's speed post-collision, focusing on how motion is transferred in the system. As forces only act momentarily during collision, the separation after collision reveals how the dynamics unfold.
Projectile Motion
Projectile motion describes the movement of objects that are thrown, projected, or launched, under the influence of gravity. While the bullet's trajectory aligns with projectile motion principles, the primary concern is its velocity after collision in a linear path.

Key aspects of projectile motion include:
  • Velocity: It consists of both horizontal and vertical components when a projectile is in motion. Here, the bullet travels horizontally into the block, and we analyze the result.
  • Gravity: It influences the vertical motion of the projectile, but not the horizontal since gravity only affects the vertical component.
  • Air resistance: Typically negligible in theoretical scenarios, it isn't considered in this problem for simplicity.
Although the bullet continues to travel after emerging from the block, this movement ties back to the calculated velocity from previous equations. This connection assures that we're on track with our conservation principles before focusing on any further projectile motion analysis. Studying how the bullet's horizontal velocity changes offers insights into the consistency and predictability of real-world physics scenarios.

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