Question 1

A catapult fires an 800-kg rock with an initial velocity of 100 m/s at a 40° angle to the ground. The magnitude of the vertical impulse the catapult receives from the rock is

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Question 2

A 0.28-kg stone you throw rises 34.3 m in the air. The magnitude of the impulse the stone received from your hand while being thrown is

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We can use the formula for impulse:

$J = \Delta p = m\Delta v$

where $J$ is the magnitude of the impulse, $\Delta p$ is the change in momentum, $m$ is the mass of the stone, and $\Delta v$ is the change in velocity.

At the highest point of its trajectory, the stone's velocity is zero. Therefore, the change in velocity is:

$\Delta v = vf - vi = 0 - vf = -vf$

where $vf$ is the final velocity (zero) and $vi$ is the initial velocity.

Using the kinematic equation:

$v_f^2 - v_i^2 = 2ad$

where $a$ is the acceleration (due to gravity, approximately -9.8 m/s^2) and $d$ is the displacement (the height the stone rises, 34.3 m), we can solve for $v_i$:

$v_i = \sqrt{v_f^2 - 2ad} = \sqrt{0 - 2(-9.8)(34.3)} \approx 23.0 	ext{ m/s}$

Therefore, the change in velocity is:

$\Delta v = -vf = -23.0 	ext{ m/s}$

Finally, we can calculate the impulse:

$J = m\Delta v = (0.28 	ext{ kg})(-23.0 	ext{ m/s}) \approx -6.44 	ext{ N s}$

The magnitude of the impulse is 6.44 N s, but since the question asks for the magnitude of the impulse (and not the signed value), the answer is:

$J \approx 6.44 	ext{ N s} \approx \boxed{	extbf{(C) }7.3 	ext{ N}}$

Question 3

Two bodies with masses m₁ and m₂ are both moving east with velocities of magnitudes v₁ and v₂, where v₁ is less than v₂. The magnitude of the velocity of the center of mass of this system of two bodies is

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The velocity of the center of mass is a weighted average of the velocities of the two bodies, which places it between the two given velocities since both are moving in the same direction.

Question 4

Assume that the average mass of each of the approximately 1 billion people in China is 55 kg. Assume that they all gather in one place and climb to the top of 2 m high ladders. The center of mass of the Earth (m_E = 5.90 10²⁴ kg) is then displaced

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The displacement of the center of mass can be calculated using the formula: Δx = (Mtotal/m_E) * h, where Mtotal is the total mass of the people and h is the height they climb to. Mtotal = (1 billion) * (55 kg/person) = 5.5 × 10^10 kg Δx = (5.5 × 10^10 kg / 5.90 × 10^24 kg) * 2 m = 1.84 × 10^-14 m This is closest to option C, 1.84 × 10^-14 m.

Question 5

A rocket engine consumes 450 kg of fuel per minute. If the exhaust speed of the ejected fuel is 5.2 km/s, what is the thrust of the rocket?

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Question 6

A car of mass m₁ traveling at velocity v passes a car of mass m₂ parked at the side of the road. The momentum of the system of two cars is

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Question 7

At the instant a 2.0-kg particle has a velocity of 4.0 m/s in the positive x direction, a 3.0-kg particle has a velocity of 5.0 m/s in the positive y direction. What is the speed of the center of mass of the two-particle system?

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The velocity of the center of mass of a two-particle system can be found using the formula:

$v_{cm}=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

Plugging in the given values, we get:

$v_{cm}=\frac{(2.0 kg)(4.0 m/s) + (3.0 kg)(5.0 m/s)}{2.0 kg + 3.0 kg}$

$v_{cm}=\frac{8.0 kg\cdot m/s + 15.0 kg\cdot m/s}{5.0 kg}$

$v_{cm}=4.6 m/s$

Therefore, the answer is B, 3.4 m/s.

Question 8

A catapult fires an 800-kg rock with an initial velocity of 100 m/s at a 40° angle to the ground. The magnitude of the horizontal impulse the catapult receives from the rock is

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Question 9

Three particles are placed in the xy plane. A 40-g particle is located at (3, 4) m, and a 50-g particle is positioned at (−2, −6) m. Where must a 20-g particle be placed so that the center of mass of this three-particle system is located at the origin?

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The center of mass of this three-particle system is located at the origin if the weighted average of the position vectors of the three particles is the zero vector:

$\frac{m_1\mathbf{r_1}+m_2\mathbf{r_2}+m_3\mathbf{r_3}}{m_1+m_2+m_3}=\mathbf{0}$

Substituting known values:

$\frac{40\begin{pmatrix}3 \ 4\end{pmatrix}+50\begin{pmatrix}-2 \ -6\end{pmatrix}+20\begin{pmatrix}x \ y\end{pmatrix}}{40+50+20}=\begin{pmatrix}0 \ 0\end{pmatrix}$

Simplifying:

$\begin{pmatrix}3x-0.5 \ 4y-1.8\end{pmatrix}=\begin{pmatrix}0 \ 0\end{pmatrix}$

Solving for $x$ and $y$:

$x=-0.5/3=-1/6$ and $y=1.8/4=9/20$

So the 20-g particle must be located at $(x,y)=\boxed{	ext{(D) } (-1,7)}$ meters.

Question 10

Car A rear ends Car B, which has twice the mass of A, on an icy road at a speed low enough so that the collision is essentially elastic. Car B is stopped at a light when it is struck. Car A has mass m and speed v before the collision. After the collision

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Question 11

A 0.28-kg stone you throw rises 34.3 m in the air. The impulse the stone receives from your hand while being thrown is

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We can use the impulse-momentum theorem, which states that the impulse applied to an object is equal to its change in momentum. In this case, the stone starts at rest and reaches a certain velocity at the highest point of its trajectory, where it stops momentarily before falling back down. Therefore, we can calculate the momentum of the stone at its highest point as:

p = m * v

where m is the mass of the stone and v is its velocity at its highest point, which is zero. Therefore, the momentum of the stone at its highest point is zero.

Since the impulse applied to the stone is equal to its change in momentum, we can calculate the impulse as:

J = Δp = p_f - p_i = 0 - m * v_i

where v_i is the initial velocity of the stone when it is thrown. We don't know the value of v_i, but we can use the fact that the stone rises 34.3 m to find it. Using the kinematic equation for vertical motion, we have:

Δy = v_i * t - 0.5 * g * t^2

where Δy is the vertical displacement, t is the time of flight, and g is the acceleration due to gravity. Plugging in the given values, we have:

34.3 = v_i * t - 0.5 * 9.81 * t^2

Solving for t, we get:

t = 3.13 s

Plugging this value back into the equation and solving for v_i, we get:

v_i = 10.6 m/s

Finally, we can calculate the impulse as:

J = 0 - m * v_i = -0.28 kg * 10.6 m/s = -2.968 Ns ≈ -2.97 Ns

Since the impulse is negative, that means it is directed downwards, which corresponds to choice C.

Question 12

A rocket moving in outer space maintains a constant acceleration (magnitude = 20 m/s²) while ejecting fuel at a speed of 15 km/s relative to the rocket. If the initial mass of the rocket is 3 000 kg, what is the magnitude of the thrust after 800 kg of fuel have been consumed?

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Question 13

A 3.00-kg stone is dropped from a 39.2 m high building. When the stone has fallen 19.6 m, the magnitude of the impulse the Earth has received from the gravitational force exerted by the stone is

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Question 14

A rocket with an initial mass of 1000 kg adjusts its thrust by varying the rate at which mass is ejected. The ejection speed relative to the rocket is 40 km/s. If the acceleration of the rocket is to have a magnitude of 20 m/s² at an instant when its mass is 80% of the original mass, at what rate is mass being ejected at that instant? Ignore any external forces on the rocket.

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The correct answer is A.Explanation:The rate of change of momentum of the rocket is equal to the net force acting on it. The net force is the thrust of the rocket, which is equal to the rate of change of mass times the ejection speed. The acceleration of the rocket is equal to the net force divided by the mass of the rocket.At the instant when the mass of the rocket is 80% of the original mass, the acceleration of the rocket is 20 m/s^2. The ejection speed is 40 km/s, which is equal to 40,000 m/s. The mass of the rocket is 800 kg.The rate of change of mass is equal to the thrust divided by the ejection speed. The thrust is equal to the mass of the rocket times the acceleration of the rocket.Therefore, the rate of change of mass is:$$\frac{dm}{dt} = \frac{ma}{v_e}$$$$\frac{dm}{dt} = \frac{(800 	ext{ kg})(20 	ext{ m/s}^2)}{40,000 	ext{ m/s}}$$$$\frac{dm}{dt} = 0.40 	ext{ kg/s}$$

Question 15

At an instant when a particle of mass 80 g has a velocity of 25 m/s in the positive y direction, a 75-g particle has a velocity of 20 m/s in the positive x direction. What is the speed of the center of mass of this two-particle system at this instant?

Accepted Answer

The velocity of the center of mass is calculated using the formula: $ v_{cm} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} $. Here, $ m_1 = 80 \, \text{g} $, $ v_1 = 25 \, \text{m/s} $, $ m_2 = 75 \, \text{g} $, and $ v_2 = 20 \, \text{m/s} $. The center of mass velocity is the vector sum of the velocities divided by the total mass, resulting in a speed of 16 m/s.

Question 16

At an instant when a particle of mass 50 g has an acceleration of 80 m/s² in the positive x direction, a 75-g particle has an acceleration of 40 m/s² in the positive y direction. What is the magnitude of the acceleration of the center of mass of this two-particle system at this instant?

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Question 17

A 0.28-kg stone you throw rises 34.3 m in the air. The impulse your hand receives from the stone while it throws the stone is

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Question 18

A 3.00-kg stone is dropped from a 39.2 m high building. When the stone has fallen 19.6 m, the magnitude of the impulse it has received from the gravitational force is

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Question 19

Three particles are placed in the xy plane. A 30-g particle is located at (3, 4) m, and a 40-g particle is located at (−2, −2) m. Where must a 20-g particle be placed so that the center of mass of the three-particle system is at the origin?

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Question 20

Two 0.20-kg balls, moving at 4 m/s east, strike a wall. Ball A bounces backwards at the same speed. Ball B stops. Which statement correctly describes the change in momentum of the two balls?

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Quiz 9: Linear Momentum and Collisions