Ch 11: Impulse and Momentum

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796당신이 사용하는 게 아니라요?교과서 변경

Knight Calc 5th Edition

Ch 11: Impulse and Momentum

문제 71b

11장, 문제 71b

A white ball traveling at 2.0m/s hits an equal-mass red ball at rest. The white ball is deflected by 25° and slowed to 1.5m/s. What percentage of the initial mechanical energy is lost in the collision?

검증된 단계별 안내

Step 1: Start by calculating the initial kinetic energy of the system. The white ball is moving at 2.0 m/s, and the red ball is at rest. Use the formula for kinetic energy: $ KE = \frac{1}{2} m v^2 $. Since the red ball is at rest, its initial kinetic energy is zero. The total initial kinetic energy is therefore $ KE_{initial} = \frac{1}{2} m (2.0)^2 $.

Step 2: After the collision, the white ball is deflected by 25° and slowed to 1.5 m/s. Calculate the final kinetic energy of the white ball using the same formula: $ KE_{white, final} = \frac{1}{2} m (1.5)^2 $.

Step 3: The red ball, initially at rest, will now have some velocity after the collision. Let the velocity of the red ball after the collision be $ v_{red, final} $. Use the principle of conservation of momentum to find $ v_{red, final} $. Break the momentum into x and y components: $ m v_{white, initial} = m v_{white, final} \cos(25°) + m v_{red, final, x} $ for the x-direction, and $ 0 = m v_{white, final} \sin(25°) - m v_{red, final, y} $ for the y-direction.

Step 4: Solve the momentum equations to find the magnitude of $ v_{red, final} $, which is $ \sqrt{v_{red, final, x}^2 + v_{red, final, y}^2} $. Once you have $ v_{red, final} $, calculate the final kinetic energy of the red ball: $ KE_{red, final} = \frac{1}{2} m v_{red, final}^2 $.

Step 5: Add the final kinetic energies of both balls to find the total final kinetic energy: $ KE_{final} = KE_{white, final} + KE_{red, final} $. To find the percentage of mechanical energy lost, use the formula: $ \text{Percentage lost} = \frac{KE_{initial} - KE_{final}}{KE_{initial}} \times 100 $.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.

영상 길이:

20m

도움이 되었나요?

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Conservation of Momentum

In a closed system, the total momentum before a collision is equal to the total momentum after the collision. This principle is crucial for analyzing collisions, as it allows us to set up equations based on the initial and final velocities of the objects involved. In this scenario, we can use the masses and velocities of the white and red balls to determine the final velocity of the red ball.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion, calculated using the formula KE = 0.5 * m * v², where m is mass and v is velocity. In this problem, we need to calculate the initial and final kinetic energies of both balls to determine the energy lost during the collision. Understanding how kinetic energy changes in elastic and inelastic collisions is essential for this analysis.

Energy Loss in Collisions

In inelastic collisions, some kinetic energy is transformed into other forms of energy, such as heat or sound, leading to a loss of mechanical energy. To find the percentage of energy lost, we compare the initial total kinetic energy to the final total kinetic energy after the collision. This concept is vital for understanding the efficiency of the collision and the extent of energy transformation that occurs.

관련 실천

교과서 질문

In Problems $76 comma 77 comma 78 comma$ and $79$ you are given the equation(s) used to solve a problem. For each of these, you are to write a realistic problem for which this is the correct equation(s).

$\frac{1}{2} (0.30 kg) (0 m/s)^{2} + \frac{1}{2} (3.0 N/m) (Δ x_{2})^{2} = \frac{1}{2} (0.30 kg) (v_{1 x})^{2} + \frac{1}{2} (3.0 N/m) (0 m)^{2}$

793

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교과서 질문

In Problems $76, 77, 78,$ and $79$ you are given the equation(s) used to solve a problem. For each of these, you are to finish the solution of the problem, including a pictorial representation.

$$\frac{1}{2}$ (0.30 $\text{ kg}$) (0 $\text{ m/s}$)^2 + $\frac{1}{2}$ (3.0 $\text{ N/m}$) ($\Delta$ x_2)^2 = $\frac{1}{2}$ (0.30 $\text{ kg}$) (v_{1x})^2 + $\frac{1}{2}$ (3.0 $\text{ N/m}$) (0 $\text{ m}$)^2$

557

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교과서 질문

A 45 g projectile explodes into three pieces: a 20 g piece with velocity 25 î m/s, a 15 g piece with velocity −10 î + 10ĵ m/s, and a 10 g piece with velocity −15 î − 20ĵ m/s. What was the projectile's velocity just before the explosion?

2383

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교과서 질문

A 2100 kg truck is traveling east through an intersection at 2.0 m/s when it is hit simultaneously from the side and the rear. (Some people have all the luck!) One car is a 1200 kg compact traveling north at 5.0 m/s. The other is a 1500 kg midsize traveling east at 10 m/s. The three vehicles become entangled and slide as one body. What are their speed and direction just after the collision?

The nucleus of the polonium isotope ²¹⁴Po (mass 214 u) is radioactive and decays by emitting an alpha particle (a helium nucleus with mass 4 u). Laboratory experiments measure the speed of the alpha particle to be 1.92×10⁷ m/s . Assuming the polonium nucleus was initially at rest, what is the recoil speed of the nucleus that remains after the decay?

1765

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교과서 질문

A neutron is an electrically neutral subatomic particle with a mass just slightly greater than that of a proton. A free neutron is radioactive and decays after a few minutes into other subatomic particles. In one experiment, a neutron at rest was observed to decay into a proton (mass 1.67×10^-27 kg) and an electron (mass 9.11×10^-31 kg) . The proton and electron were shot out back-to-back. The proton speed was measured to be 1.0 ×10⁵ m/s, and the electron speed was 3.0×10⁷ m/s. No other decay products were detected. How much momentum did this neutrino 'carry away' with it?

2150

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