Question 1

A small bomb, of mass 10. kg, is moving toward the North with a velocity of 5.0 m/s. It explodes into three fragments: a 5.0 kg fragment moving West with a speed of 8.0 m/s; a 4.0 kg fragment moving East with a speed of 10. m/s; and a third fragment with a mass of 1.0 kg. What is the velocity of the third fragment? (Neglect air friction.)

Accepted Answer

The answer of A small bomb, of mass 10. kg,...

Question 2

Car A (mass = 1000 kg) moves to the right along a level, straight road at a speed of 6.0 m/s. It collides directly with car B (mass = 200 kg) in a completely inelastic collision. What is the momentum after the collision if car B was initially at rest?

Accepted Answer

The initial momentum of the system is the momentum of car A, which is 1000 kg * 6.0 m/s = 6000 kg*m/s (or 6.0 KN-s) to the right. In a completely inelastic collision, momentum is conserved, so the momentum after the collision is also 6.0 KN-s to the right.

Question 3

Car A (mass = 1000 kg) moves to the right along a level, straight road at a speed of 6.0 m/s. It collides directly with car B (mass = 200 kg) in a completely inelastic collision. What is the speed of the car immediately after the collision?

Accepted Answer

In a completely inelastic collision, the two objects stick together after the collision. The conservation of momentum principle applies, which states that the total momentum before the collision equals the total momentum after the collision. The initial momentum is the sum of the momenta of the two cars: (1000 kg * 6.0 m/s) + (200 kg * 0 m/s) = 6000 kg*m/s (since car B is stationary, its initial momentum is 0). The combined mass after the collision is 1000 kg + 200 kg = 1200 kg. To find the final speed (v), use the formula for momentum (p = mv), so 6000 kg*m/s = 1200 kg * v. Solving for v gives v = 6000 kg*m/s / 1200 kg = 5.0 m/s.

Question 4

A 2.0 kg mass moving to the East at a speed of 4.0 m/s collides head-on in an inelastic collision with a stationary 2.0 kg mass. How much kinetic energy is lost during this collision?

Accepted Answer

In an inelastic collision, kinetic energy is not conserved. Initially, the kinetic energy is $ \frac{1}{2} m v^2 = \frac{1}{2} \cdot 2.0 \cdot (4.0)^2 = 16 $ J. After the collision, the two masses stick together and move with a common velocity $ v' $, which can be found using conservation of momentum: $ m_1v_1 + m_2v_2 = (m_1 + m_2)v' $. Substituting the given values, $ 2.0 \cdot 4.0 + 0 = (2.0 + 2.0) v' $, we find $ v' = 2.0 $ m/s. The final kinetic energy is $ \frac{1}{2} (2.0 + 2.0) (2.0)^2 = 8 $ J. The loss in kinetic energy is $ 16 - 8 = 8.0 $ J.

Question 5

When a light beach ball rolling with a speed of 6 m/s collides with a heavy exercise ball at rest, the beach ball's speed after the collision will be approximately

Accepted Answer

The beach ball, being much lighter than the exercise ball, will essentially bounce back with a speed very close to its original speed due to the conservation of momentum and energy, assuming an elastic collision. The heavy exercise ball will absorb very little of the beach ball's kinetic energy due to its much greater mass.

Question 6

A 30. kg child stands at one end of a floating 20. kg canoe that is 10. ft long. The child walks to the other end of the canoe. How far does the canoe move in the water assuming water friction is negligible?

Accepted Answer

The answer of A 30. kg child stands at one...

Question 7

A right isosceles triangle is cut out from a piece of sheet metal. The two equal sides are 1.0 m long each. How far is the center of gravity from the right angle (90. °) vertex of the triangle?

Accepted Answer

The center of gravity (centroid) of a triangle is located at a point where the three medians intersect. For any triangle, this point is located at a distance of 1/3 the length of the median from each vertex. In a right isosceles triangle, the median from the right angle vertex to the hypotenuse bisects the right angle, creating two 45-45-90 triangles. The length of this median can be found using the Pythagorean theorem, where the median (m) is the hypotenuse of these smaller triangles, and the legs are each 1/2 the length of the hypotenuse of the original triangle. Thus, $m = \sqrt{(0.5)^2 + (0.5)^2} = \sqrt{0.25 + 0.25} = \sqrt{0.5} = 0.7071$ meters. The centroid is 1/3 of the way along this median from the right angle, so the distance is $0.7071 \times \frac{1}{3} \approx 0.2357$ meters, which rounds to 0.24 meters, making the correct choice based on the options provided actually A, not C. My initial statement of choice C was incorrect based on the calculation provided. The correct answer, based on the calculation, should be A, 0.24 meters.

Question 8

A 3 kg mass is positioned at (0, 8), and a 1 kg mass is positioned at (12, 0). What are the coordinates of a 4 kg mass which will result in the center of mass of the system of three masses being located at the origin (0, 0)?

Accepted Answer

The center of mass of the system of three masses is given by the formula:$$(x_c, y_c) = \frac{m_1(x_1, y_1) + m_2(x_2, y_2) + m_3(x_3, y_3)}{m_1 + m_2 + m_3}$$where $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ are the coordinates of the three masses, and $m_1$, $m_2$, and $m_3$ are their respective masses.In this case, we have:$$(x_c, y_c) = \frac{3 kg(0, 8) + 1 kg(12, 0) + 4 kg(x_3, y_3)}{3 kg + 1 kg + 4 kg}$$$$(x_c, y_c) = \frac{(0, 24) + (12, 0) + (4x_3, 4y_3)}{8 kg}$$$$(x_c, y_c) = \frac{(12 + 4x_3, 24 + 4y_3)}{8}$$In order for the center of mass to be located at the origin, we must have:$$12 + 4x_3 = 0$$$$24 + 4y_3 = 0$$Solving these equations, we get:$$x_3 = -3$$$$y_3 = -6$$Therefore, the coordinates of the 4 kg mass are $(-3, -6)$.

Question 9

Two astronauts, of masses 60. kg and 80. kg, are initially at rest in outer space. They push each other apart. What is their separation after the heavier astronaut has moved 12. m?

Accepted Answer

By conservation of momentum, the center of mass remains stationary. The distance moved by the lighter astronaut is (80 kg / 60 kg) * 12 m = 16 m. The total separation is 12 m + 16 m = 28 m.

Question 10

Two astronauts, with masses 60. and 80. kg, are initially at rest in outer space. They push each other apart. How far has the heavier astronaut moved when the lighter astronaut has moved 12. m?

Accepted Answer

The conservation of momentum principle states that the total momentum of a closed system is constant if no external forces are acting on it. In this scenario, the system of two astronauts is closed and isolated in space, meaning their combined momentum remains zero as they push away from each other. Since momentum is the product of mass and velocity (p = mv), and both astronauts start from rest and exert equal and opposite forces on each other, their momenta are equal in magnitude but opposite in direction. Therefore, the ratio of their distances moved is inversely proportional to their masses. For the heavier astronaut (80 kg) and the lighter astronaut (60 kg) who moves 12 m, the distances they move are inversely proportional to their masses: $ \frac{60}{80} = \frac{12}{d} $, solving for $d$ gives $d = 9.0$ m.

Question 11

Astronaut Jennifer's lifeline to her spaceship comes loose and she finds herself stranded, "floating" 100. m from the mothership. She throws her 2.00 kg wrench at 20.0 m/s in a direction away from the ship. If she and her spacesuit have a combined mass of 200. kg how long does it take her to coast back to her spaceship?

Accepted Answer

To solve this, we use the conservation of momentum principle. The momentum before and after Jennifer throws the wrench must be equal because there are no external forces acting on the system. The momentum of the wrench is given by $p = mv = 2.00\, \text{kg} \times 20.0\, \text{m/s} = 40.0\, \text{kg}\cdot\text{m/s}$. Since momentum is conserved, Jennifer's momentum in the opposite direction will also be $40.0\, \text{kg}\cdot\text{m/s}$. Her velocity $v$ can be found using $p = mv$, so $v = \frac{p}{m} = \frac{40.0\, \text{kg}\cdot\text{m/s}}{200.0\, \text{kg}} = 0.2\, \text{m/s}$. To find the time $t$ it takes to travel back to the spaceship, we use $t = \frac{d}{v} = \frac{100.0\, \text{m}}{0.2\, \text{m/s}} = 500\, \text{s}$.

Quiz 6: Linear Momentum and Collisions