Ch 11: Impulse and Momentum

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 11: Impulse and Momentum

Problem 81

Chapter 11, Problem 81

A 1000 kg cart is rolling to the right at 5.0 m/s. A 70 kg man is standing on the right end of the cart. What is the speed of the cart if the man suddenly starts running to the left with a speed of 10 m/s relative to the cart?

Verified step by step guidance

Identify the principle to use: This is a conservation of momentum problem. The total momentum of the system (cart + man) before and after the man starts running must remain the same because no external horizontal forces are acting on the system.

Define the variables: Let the mass of the cart be $ m_c = 1000 \; \text{kg} $, the mass of the man be $ m_m = 70 \; \text{kg} $, the initial velocity of the cart and man (together) be $ v_i = 5.0 \; \text{m/s} $, and the velocity of the man relative to the cart be $ v_{m/c} = -10 \; \text{m/s} $ (negative because the man runs to the left). Let the final velocity of the cart be $ v_f $.

Express the man's velocity relative to the ground: The velocity of the man relative to the ground is $ v_m = v_f + v_{m/c} $. Substituting $ v_{m/c} = -10 \; \text{m/s} $, we get $ v_m = v_f - 10 \; \text{m/s} $.

Write the conservation of momentum equation: The total momentum before the man starts running is $ (m_c + m_m) v_i $. The total momentum after the man starts running is $ m_c v_f + m_m v_m $. Set these equal: $ (m_c + m_m) v_i = m_c v_f + m_m v_m $.

Substitute known values and solve for $ v_f $: Replace $ v_m $ with $ v_f - 10 \; \text{m/s} $ in the equation. This gives $ (1000 + 70)(5.0) = 1000 v_f + 70(v_f - 10) $. Simplify and solve for $ v_f $ to find the final velocity of the cart.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Conservation of Momentum

The principle of conservation of momentum states that in a closed system, the total momentum before an event must equal the total momentum after the event, provided no external forces act on it. In this scenario, the cart and the man form a closed system, and their combined momentum must remain constant as the man runs.

Relative Velocity

Relative velocity is the velocity of one object as observed from another object. In this problem, the man's speed is given relative to the cart, meaning we must account for both the cart's initial speed and the man's running speed to determine the final speed of the cart after the man starts running.

System Mass

The total mass of a system is crucial for calculating momentum. In this case, the system consists of the cart and the man. The combined mass affects how the momentum is distributed between the two when the man changes his velocity, impacting the final speed of the cart.

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Textbook Question

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).

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

Textbook Question

A 20 kg wood ball hangs from a 2.0-m-long wire. The maximum tension the wire can withstand without breaking is 400 N. A 1.0 kg projectile traveling horizontally hits and embeds itself in the wood ball. What is the greatest speed this projectile can have without causing the wire to break?

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Textbook Question

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$

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Textbook Question

A rocket with a total mass of 330,000 kg when fully loaded burns all 280,000 kg of fuel in 250 s. The engines generate 4.1 MN of thrust. What is this rocket's speed at the instant all the fuel has been burned if it is launched in deep space? If it is launched vertically from the earth?

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Textbook Question

A spaceship of mass 2.0×10⁶ kg is cruising at a speed of 5.0×10⁶ m/s when the antimatter reactor fails, blowing the ship into three pieces. One section, having a mass of 5.0×10⁵ kg, is blown straight backward with a speed of 2.0×10⁶ m/s . A second piece, with mass 8.0×10⁵ kg, continues forward at 1.0×10⁶ m/s. What are the direction and speed of the third piece?

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