Physics Test 3 Practice Test

A compressible spring whose stiffness is $1000$ N/m has a relaxed length of $0.56 m$ . If the length of the spring is $0.25 m$ , what is the potential energy of the spring?

$k = 1000$ $N/m$
$r_0 = 0.56$
$r_1 = 0.25m$
$U =$ $??$

$U = \frac 12 k\Delta x²$

$U = \frac 12 \cdot 1000 \cdot (0.25 - 0.56)²$

$U = 500 \cdot (-0.31)²$

$U = 500 \cdot 0.0961$

$U = 48.05$

A package of mass $12kg$ sits on the surface of an airless asteroid mass $1.7 × 10^{14} kg$ and radius $2.0× 10³m$ . The package is launched by a large, powerful, low mass spring, whose stiffness is $315$ $N/m$ . After launch, the package is observed to be moving at a speed of 3.0 $m/s$ when it is $5 × 10^3m$ from the center of the asteroid. How much was the spring compressed to launch the package?

$m_p=12kg$
$m_A = 1.7 × 10^{14"}kg$
r_1 = 2.0 × 10³m
$k = 315$ $N/m$
$v_f = 3.0 m/s$
$r_2 = 5 × 10³m$
$s_i =$ $??$

System: Package, asteroid, spring

Surroundings: nothing

$i=$ package at surface, spring compressed

$f:$ package moving, spring relaxed

$E_f = E_i + W + Q$ ; $W=0,Q=0$

$E_f = E_i$

$E_{rest} + K_f + U_{grav,f} + U_{spring,f} = E_{rest} + K_i + U_{grav,i} + U_{spring,i}$

$\Rightarrow K_{f}+U_{grav,f}+U_{spring,f}=K_{i}+U_{grav,i}+U_{spring,i}$

$\Rightarrow \frac 12 mv_f² + -G\frac {m_1m_2}{|\overrightarrow r_f|} +\frac 12k_ss_f² = \frac 12 mv_i² + \frac {-Gm_1m_2}{|\overrightarrow r_i|} + \frac 12 k_ss_i²$

$\Rightarrow \frac 12 (12kg)(3.0m/s)²- \frac {(6.7 × 10^{-11})(12kg)(1.7 × 10^{14})}{5 × 10³m} + \frac 12 (315)(0) = \frac 12 (12kg)(0)² - \frac {(6.7 × 10^{-11})(12kg)(1.7 × 10^{14})} {2.0 × 10³} + \frac 12 (315)s_i²$

$\Rightarrow54-27.336=-68.34+157.5s_{i}^{2}$ tarrow -14.34 + 27.336 = 157.5s_i² $$ \Rightarrow s_i =0.77666 $<div data-type="horizontalRule"><hr></div>You blast off from Mars, and you turn off the rockets when you are$ 3500 km (3.5 × 10^6 m $) from the center of Mars, well above its thin atmosphere and headed away from the planet. You intended to leave Mars for good, and by the time you get very far away you want to be coasting at the speed of$ 1800 m/s $. Mars has a mass of$ 6.4 × 10^{23} $Which of the following energy graphs best describes the Mars-spacecraft system?<img src="https://knowt-user-attachments.s3.amazonaws.com/0335897b-a02a-42f9-8df6-7678fda43cd4.png" data-width="50%" data-align="center">$ r_i = 3.5 × 10^6m $$ v_f = 1800 m/s $$ M = 6.4 × 10^{23} $$ r_f = \infty $$ U_f = 0 $$ K_f = \frac 12 Mvf² $$ E_f = U_f + K_f $$ E_f = 0 + \frac 12 Mv_f² $<ul><li>This is a positive$ E_f $, so graph 4</li></ul><div data-type="horizontalRule"><hr></div>A spacecraft is in circular orbit around Earth. The total energy of the system is$ -9 × 10^{10} J. What is the minimum amount of work needed to make the spacecraft escape Earth?

E_i = -9 × 10^{10} J $to escape,$ E_f = 0 $$ \Delta E = W $$ W = E_f - E_i = 0 $$ W = 9 × 10^{10} $<div data-type="horizontalRule"><hr></div>A space probe in outer space has a mass of 108 kg, and it is travelling at a speed of 25 m/s. When it is at location$ <405,585,-350> m it begins firing two booster rockets. Rocket A exerts a constant force of <90,90,100>N until the spacecraft reaches the location <407,590,-349> m $. At this location, the speed of the rocket is 25.6 m/s.<ul><li>$ m=108kg $</li><li>$ v_i = 25 m/s $</li><li>$ r_i = <405,585,-350>m $</li><li>$ F_A = <90,90,100>N $</li><li>$ r_f = <407,590,-349>m $</li><li>$ v_f = 25.6 $</li></ul><ul><li>Q1: What is the work done by rocket A?</li></ul>$ W = F \cdot d $$ W = (90N \cdot 2m) + (90N \cdot 5m) + (100N \cdot 1m) $$ W = (180 N\cdot m) + (450 N\cdot m) + (100 N\cdot m) $$ W = 730 N\cdot m $<ul><li>Q2: What was the work done by rocket B</li></ul>Apply Energy Principle$ \Delta E = W $$ \Delta E_{rest} + \Delta K = W $$ \Rightarrow \Delta K = W $$ K_f - K_i = W $$ \frac 12 mv_f² - \frac 12 mv_i² =W $$ \frac 12 (108)(25.6)² - \frac 12 (108)(25)² = W_A $$ 35389.44 - 33750 =W_A + W_B $$ 1639.44 = W_A + W_B $$ 1639.44 = 730 + W_B $$ W_B = 909.44

A skydiver jumps out of an airplane at a high altitude and reaches a constant “terminal” speed. Which of the following statements about the falling skydiver at terminal speed is FALSE?

(a) The thermal energy of the skydiver-Earth-atmosphere system is increasing.

(b) The gravitational potential energy of the skydiver-Earth system is decreasing.

A baseball of mass 0.025 kg is released at rest from a location 3.8m above the ground. What is its speed at the instant it hits the ground?

m=0.25kg $</li><li>$ r_i = <0,3.8,0>m $</li><li>$ r_f = <0,0,0> m $</li><li>$ v_i = 0 $</li><li>$ v_f = ?? $</li></ul>$ E_f = E_i + W + Q $$ E_f = E_i $$ K_f + U_{grav,f} = K_i + U_{grav,i} $$ \frac 12 mv_f² + mg y_f = \frac 12 mv_i² + mg y_i $$ \frac 12 (0.25kg) v_f² + (0.25kg)(9.81)(0) = \frac 12 (0.25kg)(0)² + (0.25kg)(9.81)(3.8m) $$ \frac 14 v_f² = 18.62 $$ v_f² = 74.48 $$ v_f = 8.6302 $<div data-type="horizontalRule"><hr></div>A low-mass compressible spring whose stiffness is 3500 N/m is used to launch a 4 kg block straight upward. (The block is not attached to the spring!) The spring is initially compressed 0.2m, and the block is initially at rest. When the block is 1.5 m above its starting position, what is its speed?<ul><li>$ k=3500 N/m $</li><li>$ m_{block} = 4kg $</li><li>$ x_0 = -0.2m $</li><li>$ v_{block,i} = 0 m/s $</li><li>$ r_i = 0m $</li><li>$ r_f = 1.5m $</li><li>$ v_{block,f} = ??

E_f = E_i + Q + W $$ E_f = E_i $$ K_{f} + U_{grav,f} + U_{spring,f}= K_{i} + U_{grav,i} + U_{spring,i} $$ \frac 12 mv_f² + mgy_f + \frac 12kx_f² = \frac 12 mv_i² + mgy_i +\frac 12kx_0² $$ \frac 12 (4kg)v_{block,f}² + (4kg)(9.81)(1.5m) + \frac 12 (3500 N/m)(0) = \frac 12 (4kg)(0m/s) + (4kg)(9.81)(0m) + \frac 12 (3500 N/m)(0.04m) $$ \Rightarrow 2kg\cdot v_f² + 58.86 = 70 J $$ v_f² = 5.57 $$ v_f = 2.36 $<div data-type="horizontalRule"><hr></div>You put 750 grams of water in a pot and place the pot on a hot burner. You then stir the water vigorously and do 6500 joules of work. During this short time the water’s temperature increases from$ 19\degree C to 24\degree C $. How much energy was transferred into the water due to a temperature difference between the water and the burner? In the question user 4.2 J/(gram K) for the specific heat capacity of water<ul><li>$ m=750g $</li><li>$ W = 6500 $</li><li>$ \Delta T = 5\degree C $</li></ul>$ E_f = E_i + W + Q $$ \Delta E_{thermal} = W + Q $$ mC\Delta T= 6500 + Q $$ (750)\cdot (4.2)\cdot (5) = 6500 + Q $$ 15750-6500 = Q $$ Q = 9250 J $<div data-type="horizontalRule"><hr></div>A box of initial temperature$ 20 \degree C $, mass 0.5 kg, and specific heat 600 J/K/kg is pulled horizontally across a very hot metal surface at constant speed. The rope pulling the box does 100 J of work. During the process, the box’s temperature rises to$ 21 \degree C $. What is the change in thermal energy of the box?<ul><li>$ m=0.5kg $</li><li>$ \Delta T = 1 \degree C $</li><li>$ C = 600 J/K/kg $</li><li>$ W = 100 J $</li></ul>$ E_f = E_i + W + Q $$ \Delta E = W + Q $ $ \Delta E_{thermal} = W + Q $$ mC\Delta T = (0.5kg) \cdot (600 J/K/kg) \cdot (1\degree C) = 300 J $The answer is 300 J since we are finding$ \Delta E $, not$ Q $<div data-type="horizontalRule"><hr></div>A 0.5 gram raindrop falls at its terminal speed for 100m. What is the change in thermal energy of the raindrop and surrounding air during the process?$ m = 0.5 g $$ v_{term,y,i} = v_i $$ r_f= -100m $$ \Delta E_{thermal} = ?? $System: raindrop, air, EarthSurroundings: Nothing$ \Delta K + U_{grav} + \Delta E_{thermal}=0 $$ U_{grav} + \Delta E_{thermal} =0 $$ \Delta E_{thermal} = -U_{grav} $$ -U_{grav} = -mg\Delta y = -(0.5g)(9.81N)(-100m) = -490.5g N m $$ 490.5 × 10^{-3} = 0.49 J $For the raindrop in the previous question, suppose that the air does not gain thermal energy, and all of the change in thermal energy of the change in thermal energy of the raindrop-air system goes into the rain-drop. (This is an inaccurate assumption, but it gives us a maximum estimate.) What is the change in temperature of the raindrop in this case? Water has a specific heat capacity of 4190$ J/(kg\cdot K) $.$ \Delta E_{thermal} = mC\Delta T

0.49 J= (0.0005kg)(4190 J/(kg\cdot K))\Delta T $$ \Delta T = 0.234 \degree C $<div data-type="horizontalRule"><hr></div>Ethanol has a specific heat capacity of$ 2428 J/(kg\cdot K) $. Water has a specific heat capacity of$ 4190 J/(kg\cdot K) $. If you mix a mass of ethanol at$ 10 \degree C with an equal mass m of water at a temperature of 30\degree C $, the equilibrium temperature of the mixture will be:a) less than$ 20\degree C $b) more than$ 20\degree C $c)$ 20\degree C $<ul><li>$ C_{Ethanol} = 2428 J/(kg \cdot K) $</li><li>$ C_{Water} = 4190 J/(kg\cdot K) $</li></ul><ul><li>$ \Delta E_{sys} = 0 $</li><li>$ \Delta E_{Ethanol} = \Delta E_{Water} $</li></ul><ul><li>$ mC_{Ethanol} \Delta T_{Ethanol} = -mC_{Water}\Delta T_{Water} $</li><li>$ \Rightarrow C_{Ethanol} \Delta T_{Ethanol} = -C_{Water} \Delta T_{Water} $</li></ul><div data-type="horizontalRule"><hr></div>You have a certain mass$ m of aluminum. It takes 1000 J to raise its temperature form 25\degree C to 50 \degree C $. Increasing its temperature from$ 50 \degree C to 75\degree C $, will require:<ul><li>a) less than 1000 J</li><li>b) more than 1000 J</li><li>c) the same amount of energy, 1000 J</li></ul>$ \Delta E_{thermal} = mC\Delta T $They both have the same difference.<div data-type="horizontalRule"><hr></div>A positron is a particle that has the same mass but opposite charge of an electron. A positron and an electron are initially very far apart and travel toward each other with a certain initial speed. Define the system to be the positron and electron.<ul><li>Q1 As the distance between them decreases , the electric (Coulomb) potential energy of the system:<ul><li>Decreases</li><li>Increases</li><li>Remains constant</li></ul></li></ul><ul><li>Q2 As the distance between them decreases, the kinetic energy of the system<ul><li>Decreases</li><li>Increases</li><li>Remains constant</li></ul></li></ul><ul><li>Q3 When they start very far from each other, each particle has a kinetic energy of$ 4×10^{-14} J $. When they collide, they disintegrate and completely transform into two photons. What is the total (combined) energy of the two photons after the collision?</li></ul><ul><li>Initial:<ul><li>$ r = \infty $</li><li>$ U =0 $</li><li>$ K_{elec} = K_{pos} = 4 × 10^{-14} $</li></ul></li><li>Final:<ul><li>$ 2 photons

m_{photons} = 0 $</li></ul></li></ul>$ E_f = E_i $$ K_{photons} = E_{rest,elec} + E_{rest,pos} + K_{elec} + K_{pos} $$ K_{photons} = mc² + mc²+2(4 × 10^{-14} J) $$ \Rightarrow 2mc² + 8 × 10^{-14}J $$ \Rightarrow 2(9.1 × 10^{-31}) \cdot (3 × 10^8)² + 2(4 × 10^{-14}J) $$ =2.438 × 10^{-13} $<div data-type="horizontalRule"><hr></div>A force$ F_x = -cx³ acts on an object (the y and z components are zero). What is the potential energy of the system? (Assume that U=0 at x=0 since adding a constant does not affect the Energy Principle).

F_x = -cx³ $$ F_y = 0 $$ F_z=0 $$ U=?? $Step 1. Relate force and potential energy<ul><li>$ F_x = -\frac {dU}{dx} $</li></ul>Step 2. Substitute and solve for$ U(x) $<ul><li>$ -cx³=-\frac {dU}{dx} $</li><li>$ cx³dx= dU $</li><li>$ \int {cx³dx} = U $</li><li>$ U = c\int x³dx $</li><li>$ U=\frac {cx^4}4 $</li></ul><div data-type="horizontalRule"><hr></div>A 400-kg roller coaster starts at a height of 40 m and travels down a track to a height of 10m. What is the work done by Earth on the roller coaster (For convenience in this question, use g = 10 N/kg.)<ul><li>$ m=400kg $</li><li>$ r_i = 40m $</li><li>$ r_f = 10m $</li></ul><ul><li>$ F_g = 400kg × 10 N/kg $</li><li>$ F_g = 4000 kg $</li></ul><ul><li>$ W = F_g \cdot |\Delta r| $</li><li>$ W = 4000kg \cdot |(10 - 40)| = 4000kg \cdot 30 m $</li><li>$ W = 120000kg\cdot m $</li><li>$ W = 1.2 × 10^5 kg \cdot m $</li></ul><div data-type="horizontalRule"><hr></div>When a net force doubles the magnitude of the momentum of an object, its kinetic energy does what?$ K = \frac 12 mv² $$ p = mv $$ K = \frac 12 \frac {p²}{m}

2p results in 4k

Two bound neutral atoms can be modeled as masses connected by springs - a harmonic oscillator. If a molecule absorbs energy, then:

(a) its frequency increases and its amplitude remains the same.

(b) its amplitude increases and its frequency remains the same.

(d) both its amplitude and frequency remain constant.

K + U = \frac 12 kA² $$ \omega = \sqrt{\frac km} $<ul><li>If$ K+U increases, then A increases

\omega depends on k and m and doesn’t change

Questions 25-27:

In the lab, a glider on a level, horizontal, low-friction air track has a mass of 0.1kg. It is attached to a low-mass compressible spring of stiffness 15 N/m. You pull back the glider, stretching the spring 0.1 cm from its relaxed length, and you release it from rest

m=0.1kg $</li><li>$ k_s=15 N/m $</li><li>$ s = 0.1 cm = 1 × 10^{-2}km $</li></ul>a) What is K + U for the glider-spring system?<ul><li>$ K_i = \frac 12 mv² $</li><li>$ U_{spring,i} = \frac 12 k_ss² $</li></ul><ul><li>$ K_i = 0 $</li></ul><ul><li>$ U_{spring} = \frac 12 \cdot (15 N/m) \cdot (1 × 10^{-2}km)² $</li><li>$ U_{spring} = 7.5 N/m \cdot 1 × 10^{-4}km² $</li><li>$ U_{spring} = 7.5 × 10^{-4} J $</li></ul><ul><li>$ K + U = 7.5 × 10^{-4} J $</li></ul>b) What is the speed of the glider when it passes through position where the spring is at relaxed length?<ul><li>at$ s=0 $,$ U_f = 0 $</li><li>$ K_f + U_f = K_f + U_i $<ul><li>$ \Rightarrow K_f = U_i $</li></ul></li></ul><ul><li>$ \frac 12 mv² = 7.5 × 10^{-4} J $</li><li>$ \frac 12 (\frac 1{10}kg)v² = 7.5 × 10^{-4} J $</li><li>$ v² = 20 \frac 1{kg}\cdot 7.5 × 10^{-4} J $</li><li>$ v² = 1.5 × 10^{-2} \frac J{kg} $</li><li>$ v = \sqrt{1.5 × 10^{-2} \frac J{kg}} $</li><li>$ v = 0.122 m/s $</li></ul>c) When the spring is compressed 0.005 m from its relaxed length, what is the speed of the glider?<ul><li>$ s = 0.005 m $</li><li>$ v_f = ?? $</li></ul><ul><li>$ K_f + U_f = K_i + U_f = 7.5 × 10^{-4} $</li><li>$ \frac 12 mv_f² + \frac 12 k_s s² = 7.5 × 10^{-4} $</li><li>$ \frac 12 (\frac 1{10} kg)v_f² + \frac 12(15 N/m)(0.005 m)² = 7.5 × 10^{-4}

\frac 1{20} kg \cdot v²_f +7.5 N/m \cdot 0.000025m²= 7.5 × 10^{-4} $</li><li>$ \frac 1{20} kg\cdot v_f² + 0.0001875 N\cdot m = 7.5 × 10^{-4} $</li><li>$ \frac 1{20} kg \cdot v_f² = 7.5 × 10^{-4} - 0.0001875 N\cdot m $</li><li>$ \frac 1{20} kg\cdot v_f² = 0.0005625 $</li><li>$ v_f² = 20 \cdot 0.0005625 $</li><li>$ v_f = \sqrt{0.01125} $</li><li>$ v_f = 0.10607m/s $</li></ul><div data-type="horizontalRule"><hr></div>Which of the graphs below shows the potential energy of two neutral atoms bound together in a molecule?<img src="https://knowt-user-attachments.s3.amazonaws.com/496bc613-a405-4ed4-aea7-e10661fd1261.png" data-width="100%" data-align="center">They are repulsive for small r and attractive for large r<div data-type="horizontalRule"><hr></div>Questions 30-32 relate to the same scenario. The apparatus in the diagram consists of three identical 1.6 kg masses each connected to the center of the apparatus by a 0.5 m long lightweight rod. The apparatus is in outer space, and its center of mass is moving with a velocity$ \overrightarrow v = < -6.80, 0,0> m/s $, while the apparatus rotates about its center of mass at$ \omega = 12.5 rad/sec in the counter-clockwise direction.
- m = 1.6 kg $</li><li>$ d = 0.5 m $</li><li>$ \overrightarrow v_{CM}= < -6.80,0,0> m/s $</li><li>$ \omega = 12.5 rad/s (pointing counter-clockwise)
a) What is the total kinetic energy of the apparatus
- K_{total} = K_{trans} + K_{rel} $</li></ul><ul><li>$ K_{trans} = \frac 12 M|\overrightarrow v_{CM}|² $</li><li>$ K_{trans} = \frac 12 (3 × 1.6kg)\cdot |<-6.80,0,0>|² $</li><li>$ K_{trans} = 2.4kg \cdot 46.24 $</li><li>$ K_{trans} = 110.976 J
- K_{rel} = K_{rot} + K_{vib} $</li></ul><ul><li>$ K_{rot} = \frac 12 I\omega² $</li><li>$ K_{vib} = 0 $</li></ul><ul><li>$ I = m_1r_1² + m_2r_2² + …$
- I = 3\cdot (1.6 m \cdot (0.5)²) $</li></ul><ul><li>$ K_{rot} = \frac 12\cdot 1.2 \cdot (12.5 rad/s)² $</li><li>$ K_{rot} = 93.75 $</li></ul><ul><li>$ K_{total} = 93.75 + 110.976 $</li><li>$ K_{total} = 204.726 $</li></ul>b) What is the translational angular momentum of the apparatus with respect to location A in the diagram<ul><li>$ \overrightarrow L_{A} = \overrightarrow r_A × \overrightarrow v_{cm} $</li></ul>Just miss this one dudec) What is the rotational angular momentum of the apparatus with respect to its center of mass?<div data-type="horizontalRule"><hr></div>The PY 205 instructor holds a single bicycle wheel by the axle that is approximately a hollow ring of mass 2kg, radius 0.2m, and moment of inertia$ I = MR² $. The wheel rotates counterclockwise about its center of mass at a rate of 2 rev/s. While holding the axle of the rotating wheel in the air, the instructor walks at a constant speed of 1 m/s across the room<ul><li>$ m = 2kg $</li><li>$ r = 0.2m $</li><li>$ I = (2kg) \cdot (0.2m)² = 0.4 kg \cdot m² $</li><li>$ \overrightarrow v = 1 m/s $</li><li>$ \omega = 2 rev/s $</li></ul>a) What is the angular speed of the wheel in rad/s? <ul><li>$ \omega = 2 revs $</li><li>$ \omega = 2 rev/s \cdot 2\pi \frac {rad}{rev} $</li><li>$ \omega = 4\pi \frac{rad}s $</li></ul>b) For the wheel in the previous question, which is greater, its translational kinetic energy or its rotational kinetic energy?<ul><li>$ K_{trans} = \frac 12 M|v|² $</li><li>$ K_{rot} = \frac 12 I\omega² = \frac 12 MR² \cdot \omega $</li></ul>Since$ \omega is a positive number above one, K_{rot} is greater
 A system of three particles has the following properties:
 - m_1 = 2kg $</li><li>$ \overrightarrow v_1 = <16,-8,0>m/s $</li></ul><ul><li>$ m_2 = 4kg $</li><li>$ \overrightarrow v_2 = < -15, 2,0> m/s $</li></ul><ul><li>$ m_3 = 8kg $</li><li>$ \overrightarrow v_3 = [ERROR] $</li></ul><ul><li>$ \overrightarrow v_{CM} = < -2, -0.57, 4.6> m/s $</li></ul>What is the translational kinetic energy of this system?<ul><li>$ K_{trans} = \frac 12 M|\overrightarrow v_{cm}|² $</li><li>$ K_{trans} = \frac 12 (2 + 4 + 8) \cdot |< -2, -0.57, 4.6>|² $</li><li>$ K_{trans} = 7 \cdot \sqrt {2² + 0.57²+ 4.6²} $</li><li>$ K_{trans} = 7 \cdot 5.05² $</li><li>$ K_{trans} = 7 \cdot 25.5 $</li><li>$ K_{trans} = 178.4 $</li></ul><div data-type="horizontalRule"><hr></div>A person throws a rod-like boomerang into the air with a velocity of 30 m/s. It has a mass of 1.0 kg and is rotating at 62.8 radians/sec about the center of mass. The moment of inertia around the axis of rotation is 0.041 kg$ m² $. What is the rotational kinetic energy of the system around the center of mass?<ul><li>$ v_i = 30m/s $</li><li>$ m=1kg $</li><li>$ \omega_{CM} =62.8 $</li><li>$ I_{CM} = 0.041 kg \cdot m² $</li><li>$ K_{rot, CM} = ?? $</li></ul>$ K_{rot, CM} = \frac {L_{rot,CM}²}{2I} = \frac 12 I\omega_{CM}² $$ K_{rot,CM} = 0.041 kg\cdot m² \cdot 62.7² (rad/s)² $$ K_{rot,CM} = 80.85 J $<div data-type="horizontalRule"><hr></div>A thin, lightweight rod of length 2r has a small ball of mass m attached to each end at a distance r from the center. Its moment of inertia is$ I $. If the masses are moved halfway closer to the center, to a distance$ \frac 12 r $, then the moment of inertia of the system will be$ \frac I4 $<div data-type="horizontalRule"><hr></div>In outer space, far from other objects, a block of mass$ m=3 kg travelling at speed v_1 = 10 m/s in the +x direction approaches an object initially at rest that consists of two small spheres each of mass M=20 kg connected by a low-mass rod, with center-to-center distance d=0.40 m. The block is observed to bounce off the upper sphere and proceed with speed v_2 = 8ms at an angle \theta = 30 \degree to the original direction
 - m_b=3kg $</li></ul><ul><li>$ v_1 = 10m/s $</li><li>$ v_2 = 8m/s $</li></ul><ul><li>$ M_s = 20kg
 - d = 0.40m $</li></ul><ul><li>$ \theta = 30\degree $</li></ul>Just before the collision, what is the translational angular momentum of the moving block, with respect to point A, the center of the rod?<ul><li>$ \overrightarrow L_{trans, A} = \overrightarrow r_{cm} × \overrightarrow p_{tot} $</li></ul><ul><li>$ r_{cm} = \frac d2 = 0.2 $</li><li>$ p_{tot} = mv_1 $<ul><li>$ p_{tot} = 3kg × 10 \frac ms = 30 kg \cdot \frac ms
 - L_{trans} = 0.2 \cdot 30 kg = 6 $</li></ul>$ <0, 0, -6>$$