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Chapter 8: Problem 80

A Ricocheting Bullet. A \(0.100 \mathrm{~kg}\) stone rests on a frictionless, horizontal surface. A bullet of mass \(6.00 \mathrm{~g}\). traveling horizontally at \(350 \mathrm{~m} / \mathrm{s}\), strikes the stone and rebounds horizontally at right angles to its original direction with a speed of \(250 \mathrm{~m} / \mathrm{s}\). (a) Compute the magnitude and direction of the velocity of the stonc after it is struck. (b) Is the collision perfectly clastic?

Short Answer

Expert verified
After working out the details, the magnitude and direction of the initial velocity of the stone after it is struck can be found. Further, by comparing the total kinetic energy before and after the collision, we can determine whether or not the collision was perfectly elastic.

Step by step solution

01

Identify given values

The initial mass of the stone \(m_{1} = 0.1 \: kg\) and it's initially at rest \(v_{1i} = 0 \: m/s\). The mass of the bullet \(m_{2} = 0.006 \: kg\) and it's velocity \(v_{2i} = 350 \: m/s\). After the collision, the bullet's velocity \(v_{2f} = 250 \: m/s\) at right angles to its original direction.
02

Compute the momentum after collision

By the conservation of momentum principle, the momentum before and after collision should be equal in both x (original direction of the bullet's motion) and y direction (perpendicular to the bullet's original direction). Thus, for x direction, \(m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{2}v_{2f_x}\) and for y direction, \(m_{1} 0 + m_{2} 0 = m_{1} 0 + m_{2}v_{2f_y}\). Solving these equations will give the velocities \(v_{1f}\) and \(v_{2f_x}\) of the stone and the x-component of the bullet's velocity after collision respectively since \(v_{2f_y} = 250 m/s\) (velocity of bullet after collision).
03

Calculate the velocity magnitude and direction

The magnitude and direction of the velocity \(v_{1f}\) of the stone after it is struck can be found by realizing that the stone's motion will be along the direction of the momentum's resultant, which will be a combination of the bullet's momentum after collision in the x and y directions. Pythagorean theorem could be used to find the magnitude and inverse tangent of the velocity components to find the direction.
04

Check the type of collision

An elastic collision is one in which both momentum and kinetic energy are conserved. To ascertain if the collision is elastic, calculate the total kinetic energy before and after the collision. If they are equal, the collision is elastic. If not, it is inelastic.

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Key Concepts

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

Elastic Collision
An elastic collision is a special kind of collision in which both momentum and kinetic energy are conserved. Momentum is always conserved in closed systems due to Newton's third law, but for a collision to be elastic, the energy conservation must hold as well.

During an elastic collision, such as when a bullet strikes a stone and rebounds at a right angle, both linear momentum and kinetic energy are maintained. This requires that after the collision, any losses due to sound, heat, or deformation must be negligible.

In the context of the exercise:
  • If the sum of the initial kinetic energy (from bullet and any initial motion of the stone) equals the sum of the final kinetic energy (after the bullet rebounds and the stone starts moving), the collision can be considered perfectly elastic.
  • Empirically, this means that calculations should show no difference between the total kinetic energy before and after the event.
Kinetic Energy Conservation
Kinetic energy is the energy that an object possesses due to its motion. In the context of collisions, the principle of kinetic energy conservation is crucial when evaluating elastic collisions.

For a perfectly elastic collision:
  • The total kinetic energy within the system before the collision will equal the total kinetic energy after the collision.
  • This means that all kinetic energy from moving objects is transferred without any loss.
In the exercise:
  • To check if the collision was elastic, calculate the bullet鈥檚 kinetic energy before impact: \( KE_{initial} = \frac{1}{2} m_{2} v_{2i}^2 \)
  • And after the impact for both the rebounded bullet and moving stone: \( KE_{final} = \frac{1}{2} m_{2} v_{2f}^2 + \frac{1}{2} m_{1} v_{1f}^2 \)
  • If these energies are equal, the collision is elastic.
Pythagorean Theorem
The Pythagorean Theorem is a crucial tool in physics for analyzing motion in two dimensions. Typically used in problems dealing with vectors, it states that in a right-angled triangle, the square of the hypotenuse (longest side) is equal to the sum of the squares of the other two sides.
  • Mathematically, it is represented as: \( c^2 = a^2 + b^2 \), where \(c\) is the hypotenuse, and \(a\) and \(b\) are the other sides.
In relation to the exercise:
  • The bullet鈥檚 motion after collision could be split into x and y components that together form a right triangle.
  • The magnitude of the stone's velocity after collision can be calculated using the Pythagorean theorem by combining these perpendicular components of momentum.
  • This means: \( v_{resultant} = \sqrt{v_{xf}^2 + v_{yf}^2} \).
Utilizing the Pythagorean theorem in this context allows us to derive a precise resultant velocity by incorporating all directional changes post-collision.

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Most popular questions from this chapter

You are at the controls of a particle accelerator, sending a beam of \(1.50 \times 10^{7} \mathrm{~m} / \mathrm{s}\) protons (mass \(\mathrm{m}\) ) at a gas target of an unknown element. Your detector tells you that some protons bounce straight back after a collision with one of the nuclei of the unknown element. All such protons rebound with a speed of \(1.20 \times 10^{7} \mathrm{~m} / \mathrm{s}\). Assume that the initial speed of the target nucleus is negligible and the collision is elastic. (a) Find the mass of one nucleus of the unknown element. Express your answer in terms of the proton mass \(m\). (b) What is the speed of the unknown nucleus immediately after such a collision?

When cars are equipped with flexible bumpers, they will bounce off cach other during low-specd collisions, thus causing less damage, In one such accident, a \(1750 \mathrm{~kg}\) car traveling to the right at \(1.50 \mathrm{~m} / \mathrm{s}\) collides with a \(1450 \mathrm{~kg}\) car going to the left at \(1.10 \mathrm{~m} / \mathrm{s}\) Measurements show that the heavier car's speed just after the collision was \(0.250 \mathrm{~m} / \mathrm{s}\) in its original direction. Ignore any road friction during the collision. (a) What was the speed of the lighter car just after the collision? (b) Calculate the change in the combined kinetic eneryy of the two-car system during this collision.

A10.0g marble slides to the left at a speed of \(0.400 \mathrm{~m} / \mathrm{s}\) on the frictionless, horizontal surface of an icy New York sidewalk and has a head-on, clastic collision with a larger \(30.0 \mathrm{~g}\) marble sliding to the right at speed of \(0.200 \mathrm{~m} / \mathrm{s}\) (Fig. \(\mathrm{E} 8.48\) ). (a) Find the velocity of each marble (magnitude and dircction) after the collision. (since the collision is head-on, all motion is along a line. (b) Calculate the change in momentum (the momentum after the collision minus the momentum before the collision) for each marble. Compare your values for each marble. (c) Calculate the change in kinetic energy (the kinctic energy after the collision minus the kinetic energy before the collision) for each marble. Compare your values for each marble.

A block with mass \(0.500 \mathrm{~kg}\) sits at rest on a light but not long vertical spring that has spring constant \(80.0 \mathrm{~N} / \mathrm{m}\) and one end on the floor. (a) How much clastic potential cnergy is stored in the spring when the block is sitting at rest on it? (b) A second identical block is dropped onto the first from a height of \(4.00 \mathrm{~m}\) above the first block and sticks to it. What is the maximum elastic potential energy stored in the spring during the motion of the blocks after the collision? (c) What is the maximum distance the first block moves down after the second block has landed on it?

An apple with mass \(M\) is hanging at rest from the lower end of a light vertical rope. A dart of mass \(M / 4\) is shot vertically upward, strikes the bottom of the apple, and remains cmbedded in it. If the specd of the dart is \(\mathrm{c}_{0}\) just before it strikes the apple, how high docs the apple move upward because of its collision with the dart?
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