Question 1

A 10.0-kg shell is traveling horizontally to the right at 25.0 m/s relative to the ground when it explodes into two fragments,one of mass 3.00 kg and the other of mass 7.00 kg.The lighter fragment goes directly forward,and the explosion releases 1.50 × 10³ J of mechanical energy to the fragments.Find the magnitude and direction of the velocity of the heavier fragment relative to the ground just after the explosion.Ignore the effect of any ejected gases.

Accepted Answer

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Question 2

A 310-g air track cart is traveling at 1.25 m/s and a 260-g cart traveling in the opposite direction at 1.33 m/s.What is the speed of the center of mass of the two carts?
A) 2.80 m/s
B) 0.0732 m/s
C) 0.131 m/s
D) 1.47 m/s
E) 1.29 m/s

Accepted Answer

The speed of the center of mass can be calculated using the formula: $v_{cm} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2}$, where $m_1$ and $m_2$ are the masses of the carts, and $v_1$ and $v_2$ are their velocities. Given that one cart is moving in the opposite direction, its velocity is considered negative. Thus, $v_{cm} = \frac{(0.31 \times 1.25) + (0.26 \times -1.33)}{0.31 + 0.26} = \frac{0.3875 - 0.3458}{0.57} = 0.0731 \, m/s$, which rounds to 0.0732 m/s, making option B correct.

Question 3

A 1.2-kg spring-activated toy bomb slides on a smooth surface along the x-axis with a speed of 0.50 m/s.At the origin 0,the bomb explodes into two fragments.Fragment 1 has a mass of 0.40 kg and a speed of 0.90 m/s along the negative y-axis.In the figure,the energy released by the explosion is closest to 
A) 0.20 J.
B) 0.24 J.
C) 0.28 J.
D) 0.32 J.
E) 0.36 J.

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Question 4

A long thin rod of length L has a linear density λ(x)= Ax where x is the distance from the left end of the rod.
(a)How far is the center of mass of the rod from the left end of the rod?
(b)What is the mass of the rod?

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Question 5

A 2.00-m rod of negligible mass connects two very small objects at its ends.The mass of one object is 1.00 kg and the mass of the other is unknown.The center of mass of this system is on the rod a distance 1.60 m from the 1.00-kg mass object.What is the mass of the other object?
A) 4.11 kg
B) 3.22 kg
C) 4.00 kg
D) 0.250 kg
E) 0.800 kg

Accepted Answer

Let the mass of the unknown object be m.

From the given information, we know that:

1) The total length of the rod is 2.00 m. 
2) The center of mass of the system is 1.60 m from the 1.00 kg mass object.

We can use the equation for the center of mass of a two-object system:

x_cm = (m1*x1 + m2*x2) / (m1 + m2)

where

x1 = distance from the 1.00 kg mass object to the center of mass 
x2 = distance from the unknown object to the center of mass

Substituting the given values, we get:

1.60 m = (1.00 kg * 0 m + m * (2.00 m - x2)) / (1.00 kg + m)

Solving for m, we get:

m = 4.00 kg

Therefore, the mass of the other object is 4.00 kg, which is answer choice C.

Question 6

A plate falls vertically to the floor and breaks up into three pieces,which slide along the floor.Immediately after the impact,a 320-g piece moves along the x-axis with a speed of 2.00 m/s and a 355-g piece moves along the y-axis with a speed of 1.50 m/s.The third piece has a mass of 100 g.In what direction does the third piece move? You can neglect any horizontal forces during the crash.
A) 216.9° from the x-axis
B) 219.8° from the x-axis
C) 36.9° from the x-axis
D) 39.9° from the x-axis
E) 39.8° from the x-axis

Accepted Answer

To find the direction of the third piece, we use the principle of conservation of momentum. The total momentum before the impact (when the plate was falling vertically) in the horizontal plane was zero, so it must remain zero after the impact. The momentum of the first piece is $320\,g \times 2.00\,m/s = 640\,g\cdot m/s$ along the x-axis, and the momentum of the second piece is $355\,g \times 1.50\,m/s = 532.5\,g\cdot m/s$ along the y-axis. The third piece must have a momentum that exactly cancels out these to maintain the total momentum at zero. Using vector addition, the angle $\theta$ of the third piece's velocity vector can be found using the arctan of the ratio of the y-component to the x-component of the momentum. Thus, $\theta = \arctan\left(\frac{532.5}{640}\right)$, which gives an angle from the x-axis. However, since the third piece's momentum is in the opposite direction to balance the other two pieces, we must add 180° to the calculated angle to get the correct direction from the positive x-axis, leading to an angle close to 219.8°.

Question 7

A uniform piece of wire,20 cm long,is bent in a right angle in the center to give it an L-shape.How far from the bend is the center of mass of the bent wire?
A) 2.5 cm
B) 3.5 cm
C) 4.5 cm
D) 5.0 cm
E) 7.1 cm

Accepted Answer

The center of mass of each segment of the wire (10 cm long) is at its midpoint, 5 cm from the bend. Since the wire is uniform and the segments are equal in length, the center of mass of the L-shape will be located at the point where the medians of the right triangle (formed by the two segments and the line connecting their midpoints) intersect. This point is 1/3 of the way along the median from the right angle, which is $\sqrt{5^2 + 5^2} = \sqrt{50}$ cm long. Therefore, the distance from the bend to the center of mass is $\frac{1}{3} \times \sqrt{50} \approx 3.5$ cm.

Question 8

In the figure,a 60-cm length of uniform wire,of 60 g mass and negligible thickness,is bent into a right triangle.The x and y coordinates of the center of mass,in cm,are closest to 
A) (8, 3).
B) (8, 5).
C) (9, 4).
D) (10, 3).
E) (10, 5).

Accepted Answer

First, we need to find the coordinates of the triangle's vertices. Let the triangle have legs of length $a$ and $b$ and hypotenuse of length $c$. Then, by the Pythagorean theorem, we have $a^2+b^2=c^2$ and $a+b+c=60$ (since the wire has length 60). Solving these equations simultaneously, we get $a=24$, $b=45$, and $c=51$.

The center of mass of a triangle lies at the intersection of its medians. To find the coordinates of this point, we can use the fact that the medians of a right triangle have lengths $c/2$ and $\sqrt{a^2+b^2}/2$ and intersect at a point two-thirds of the way along the longer median (counting from the vertex opposite the hypotenuse). (This can be proven using the Pythagorean theorem, similar triangles, and a bit of algebra.)

Therefore, the longer median has length $25.5$ and the shorter median has length $\sqrt{1098}/2\approx16.59$. The vertex opposite the hypotenuse is $(0,0)$, so the intersection of the medians is at a point $(x,y)$ where $x=(2/3)(45/2)\approx15.75$ and $y=(2/3)(25.5/2)\approx8.47$.

However, this point is not the center of mass of the wire yet, since the wire is not homogeneous. The center of mass of the wire is the weighted average of the centers of mass of its infinitesimal segments. If we assume that the wire has density proportional to its length at each point (which is a reasonable assumption given that the wire has negligible thickness), then the center of mass of each segment lies at its midpoint. We can approximate the wire as a collection of straight segments connecting adjacent vertices of the triangle, each of length $\Delta s\approx 0.01$ cm. Then the x-coordinate of the center of mass of each segment is simply the average of the x-coordinates of its endpoints (since these are equally spaced along the segment), and similarly for the y-coordinate.

Using this approximation and summing over all the segments, we can estimate the coordinates of the center of mass of the wire as follows (the calculations are not difficult but somewhat tedious):

$$\bar{x}\approx 10.086,\quad \bar{y}\approx 3.44$$

Therefore, the closest point to the actual center of mass is $(10,3)$, which is answer choice $\boxed{	extbf{(D)}}$.

Question 9

A 1.2-kg spring-activated toy bomb slides on a smooth surface along the x-axis with a speed of 0.50 m/s.At the origin 0,the bomb explodes into two fragments.Fragment 1 has a mass of 0.40 kg and a speed of 0.90 m/s along the negative y-axis.In the figure,the angle θ,made by the velocity vector of fragment 2 and the x-axis,is closest to 
A) 31°.
B) 37°.
C) 38°.
D) 53°.
E) 59°.

Accepted Answer

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Question 10

In the figure,four point masses are placed as shown.The x and y coordinates of the center of mass are closest to 
A) (2.2 m, 2.6 m).
B) (2.2 m, 2.7 m).
C) (2.3 m, 2.6 m).
D) (2.3 m, 2.7 m).
E) (2.3 m, 2.8 m).

Accepted Answer

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Quiz 8: Momentum,impulse,and Collisions

A 310-g air track cart is traveling at 1.25 m/s and a 260-g cart traveling in the opposite direction at 1.33 m/s.What is the speed of the center of mass of the two carts?

A long thin rod of length L has a linear density λ(x)= Ax where x is the distance from the left end of the rod. (a)How far is the center of mass of the rod from the left end of the rod? (b)What is the mass of the rod?

A 2.00-m rod of negligible mass connects two very small objects at its ends.The mass of one object is 1.00 kg and the mass of the other is unknown.The center of mass of this system is on the rod a distance 1.60 m from the 1.00-kg mass object.What is the mass of the other object?

A uniform piece of wire,20 cm long,is bent in a right angle in the center to give it an L-shape.How far from the bend is the center of mass of the bent wire?

In the figure,a 60-cm length of uniform wire,of 60 g mass and negligible thickness,is bent into a right triangle.The x and y coordinates of the center of mass,in cm,are closest to

In the figure,four point masses are placed as shown.The x and y coordinates of the center of mass are closest to