Physics flashcards for exam 2- examples and steps
Geoff counts the number of oscillations of a simple pendulum at a location where the acceleration due to gravity is 9.80 m/s2 and finds that it takes 7.75 s for 13.0 complete cycles. Calculate the length 𝐿 of the pendulum. 𝐿= ____m
A simple pendulum's period depends on its length and the acceleration due to gravity and is independent of the pendulum's mass. Therefore, near the surface of Earth, the only quantity that separates one pendulum from another is the length of the connector between its axis of rotation and the pendulum bob. The longer the connector, the longer the pendulum's period. The period 𝑇 of a pendulum is the time for one complete cycle of an oscillatiton. Thus, given the time 𝑡 for a given number 𝑛 of cycles, the period is the ratio of 𝑡 to 𝑛. 𝑇=𝑡/𝑛 The period of a pendulum is determined by the pendulum's length 𝐿 and the accleration due to gravity 𝑔. 𝑇=2𝜋√𝐿/𝑔 Substitute the expression for 𝑇, then solve for 𝐿. L=(t/2𝜋n)^2g Substitute the given values and evaluate. =0.0882 m
An observer sits in a boat watching wave fronts move past the boat. The distance between successive wave crests is 0.70 m, and they are moving at 2.4 m/s. What is the wavelength of these waves? 2.4 m 1.4 m 0.70 m What is the frequency of these waves? 3.4 Hz 0.70 Hz 0.29 Hz What is the period of these waves? 3.4 s 0.70 s 0.29 s
A wave is a disturbance that travels from one location to another, usually through a medium. A wave that is periodic has a wavelength, speed, frequency, and period associated with it. The distance between the wave crests is the wavelength 𝜆. 𝜆=0.70 m The speed 𝑣 is the product of frequency 𝑓 and wavelength 𝜆, so frequency is the speed divided by the wavelength. 𝑓=𝑣/𝜆 =2.4 m/s / 0.70 m =3.4 Hz The frequency is the number of cycles per unit time. The unit of Hertz is cycles per second. Sometimes this is just shown as 1/s or s−1. The period 𝑇 varies inversely with the frequency and measures the amount of time for one complete cycle. 𝑇= 1 / 𝑓 =1 / 3.4 Hz =0.29 s
A point rotates about the origin in the 𝑥𝑦 plane at a constant radius of 0.218 m with an angular velocity of 7.45 rad/s. The projection of this point's motion on the 𝑥 - or 𝑦 -axis is simple harmonic. What is the amplitude of the projected simple harmonic motion?____m What is the projected motion's frequency? ____Hz What is the projected motion's period? _____s
Express the point's position on the circle in (𝑥,𝑦) coordinate form by parameterizing its motion in terms of the radius 𝑟, the angular momentum 𝜔, and the time 𝑡. (𝑥,𝑦)(𝑡) = (𝑟cos(𝜔𝑡),𝑟sin(𝜔𝑡)) Using this parameterization, note that the projections of the motion on the 𝑥-axis, 𝑥(𝑡), and the 𝑦-axis, 𝑦(𝑡), are simply the parameterizations of the 𝑥 and 𝑦 positions of the point on the circle. 𝑥(𝑡)=𝑟cos(𝜔𝑡) 𝑦(𝑡)=𝑟sin(𝜔𝑡) These equations for 𝑥(𝑡) and 𝑦(𝑡) are identical in form to equations describing simple harmonic motion, as they must be. Consequently, the radius 𝑟 is the amplitude 𝐴 of the motion. 𝐴=𝑟=0.218 m The frequency 𝑓 is related to the angular frequency in the usual way. 𝑓=𝜔/2𝜋 =1.19 Hz The period 𝑇 is the time required to make the argument of the sine or cosine function equal to 2𝜋. 𝑇= 2𝜋/𝜔 =0.843 s
A block is attached to a ceiling by a spring of force constant 𝑘 . When pulled down and released, the block undergoes simple harmonic motion. The motion of the block is shown in the time plot. Which of the choices most closely describes the amplitude 𝐴 and frequency 𝑓 of the motion of the block? 𝐴=4 cm ; 𝑓=1/8𝜋 Hz 𝐴=2 cm ; 𝑓=1/16𝜋 Hz 𝐴=4 cm ; 𝑓=1/16𝜋 Hz 𝐴=2 cm ; 𝑓=1/6𝜋 Hz
Finding the period of oscillation is usually the easiest way to begin analyzing sine (and cosine) plots. Here, the period 𝑇 equals 16𝜋, since that is the time needed to complete the cycle. The frequency is simply the inverse of the period, so 𝑓=116𝜋. The amplitude 𝐴 is the maximum distance of travel from the origin. Here, 𝐴=2 cm.
Calculate the momentum 𝑝elephant of a 2140 kg elephant charging a hunter at a speed of 7.59 m/s.
Given: mass and velocity Equation: p=mv Pelephant= (2140kg)(7.59 m/s)= 16200 kg*m/s
What is the momentum of a garbage truck that is 13500 kg and is moving at 10.0 m/s?
Given: mass and velocity Equation: p=mv 𝑝truck = (13500 kg)(10.0 m/s) =1.35×10^5 kg⋅m/s OR 135000
Calculate the final velocity right after a 115 kg rugby player who is initially running at 7.75 m/s collides head‑on with a padded goalpost and experiences a backward force of 18500 N for 5.50×10−2 s.
In order to find the final velocity of the football player, use the fact that the change in momentum is equal to the impulse Δ𝑝=𝑝f−𝑝i=𝐹netΔ𝑡 where Δ𝑝 is the change in momentum, 𝐹net is the net external force, and Δ𝑡 is the change in time. Also, use the definition of momentum, which states that 𝑝=𝑚𝑣 where 𝑚 is the mass and 𝑣 is the velocity. Plugging the definition of momentum into the impulse-momentum relationship gives 𝑚𝑣f−𝑚𝑣i=𝐹netΔ𝑡 Rearrange for the final velocity. 𝑣f=𝑚𝑣i+𝐹netΔ𝑡𝑚 The force on the player from the goalpost is backward or negative. Hence 𝑣f=(115 kg)(7.75 m/s)+(−18500 N)(5.50×10−2 s) / 115 kg = −1.10 m/s
A 27.0 kg shopping cart is moving with a velocity of 4.0 m/s. It strikes a 9.0 kg box that is initially at rest. They stick together and continue moving at a new velocity. Assume that friction is negligible. What is the velocity of the combined shopping cart-box wreckage after the collision? 4.0 m/s 0 m/s 12 m/s 3.0 m/s
The final step is to solve the equation for the final velocity of the combined cart-box wreckage after the collision by dividing the total momentum by the total mass of the cart and box. velocity of wreckage = total momentum / mass of wreckage =108 kg⋅m/s / 36.0 kg =3.0 m/s
A trombone has a variable length. When a musician blows into the mouthpiece and causes air in the tube of the horn to vibrate, the waves set up by the vibrations reflect back and forth in the horn to create standing waves. As the length of the horn is made shorter, what happens to the frequency? Assume that the same harmonic is excited. The frequency remains the same. The frequency increases. The frequency decreases. The frequency increases or decreases depending on how hard the horn player blows. The frequency increases or decreases depending on the diameter of the horn.
The frequency increases.
regular collisions
elastic
Identify the statements about springs as either true or false. As you compress a spring from equilibrium, the force the spring exerts on you decreases.
false
Identify the statements about springs as either true or false. As you stretch a spring from equilibrium, the force the spring exerts on you decreases.
false
Two tuning forks of frequency 480 Hz and 484 Hz are struck simultaneously. What is the beat frequency resulting from the two sound waves? 0 Hz 2 Hz 4 Hz 482 Hz 964 Hz
The beat frequency is the absolute value of the difference between the frequencies. 𝑓beats=∣∣𝑓1−𝑓2∣∣ =∣∣480 Hz−484 Hz∣∣ =4 Hz
sticky collisions
inelastic
Two violinists are trying to play in tune. However, whenever they play their A string at the same time they hear a beat frequency of 3.0 Hz. One of the violinists measures her A string with an electronic tuner and finds that it is tuned to 422 Hz. What are the possible frequencies for the A string of the other violinist? (select all that apply) 422 Hz 428 Hz 425 Hz 416 Hz 419 Hz
The beat frequency 𝑓beat is defined by the equation 𝑓beat=|𝑓2−𝑓1| where 𝑓1 and 𝑓2 are the two interfering wave frequencies. Therefore, determine the possible frequencies by adding and subtracting 𝑓beat from the known frequency. 𝑓2=𝑓1+𝑓beat =422 Hz+3.0 Hz =425 Hz 𝑓2=𝑓1−𝑓bea t=422 Hz−3.0 Hz =419 Hz The possible frequencies are 425 Hz and 419 Hz.
If the amplitude of a sound wave is tripled, the intensity will increase by a factor of 9 . remain the same. decrease by a factor of 3 . increase by a factor of 3 . decrease by a factor of 9 .
The energy carried by a wave is proportional to the square of the amplitude. The power is the energy per unit time, and the intensity of a wave is the power per unit area, 𝐸∝𝐴^2 Therefore, tripling the amplitude of a wave results in the intensity increasing by a factor of 9.
Which of the equations of motion best describes the motion of the block? 𝑦(𝑡)=4cos(𝑡8) 𝑦(𝑡)=4cos(16𝜋𝑡) 𝑦(𝑡)=2cos(𝑡8) 𝑦(𝑡)=2cos(8𝑡)
The general form of the equation of motion for simple harmonic motion is 𝑦(𝑡)=𝐴cos(𝜔𝑡+𝜙) In this case, the phase 𝜙=0 since the block begins at its maximum positive displacement. The angular frequency 𝜔=18 s−1 since 𝜔=2𝜋𝑓. Substitute the values into the equation of motion. 𝑦(𝑡)=2cos(𝑡 / 8 s) For simple harmonic motion, the relationship between the force constant 𝑘 and the mass 𝑚 of the block is given by 𝜔=√𝑘/𝑚 Solve for the force constant and evaluate. 𝑘=𝑚𝜔^2 =8.14 kg×(1/8 s−1)^2 =0.127 N/m
A 27.0 kg shopping cart is moving with a velocity of 4.0 m/s. It strikes a 9.0 kg box that is initially at rest. They stick together and continue moving at a new velocity. Assume that friction is negligible. What was the momentum of the box before the collision? 216 kg⋅m/s 0 kg⋅m/s 36 kg⋅m/s 108 kg⋅m/s
momentum of box=mass of box×velocity of box =9.0 kg×0 m/s =0 kg⋅m/s
A 27.0 kg shopping cart is moving with a velocity of 4.0 m/s. It strikes a 9.0 kg box that is initially at rest. They stick together and continue moving at a new velocity. Assume that friction is negligible. What was the momentum of the shopping cart before the collision? 0 kg⋅m/s 216 kg⋅m/s 108 kg⋅m/s 36 kg⋅m/s
momentum of cart = mass of cart × velocity of cart =27.0 kg×4.00 m/s =108 kg⋅m/s
A 1.35 kg falcon catches a 0.335 kg dove from behind in midair. What is their velocity after impact if the falcon's velocity is initially 27.5 m/s and the dove's velocity is 6.95 m/s in the same direction? velocity:______m/s
The equation for momentum 𝑝 is 𝑝=𝑚𝑣 where 𝑚 is the mass of the object and 𝑣 is the object's velocity. The initial momentum is 𝑝i = 𝑚falcon𝑣i,falcon + 𝑚dove𝑣i,dove And the final momentum is 𝑝f = (𝑚falcon+𝑚dove)𝑣f Using conservation of momentum, set the initial momentum equal to the final momentum. 𝑚falcon𝑣i,falcon + 𝑚dove𝑣i,dove=(𝑚falcon + 𝑚dove)𝑣f Next, solve for the final velocity. 𝑣f = 𝑚falcon𝑣i,falcon + 𝑚dove𝑣i,dove / 𝑚falcon+𝑚dove Finally, plugging in the numbers gives 𝑣f = (1.35 kg)(27.5m/s)+(0.335 kg)(6.95m/s) / 1.35 kg+0.335 kg =23.4m/s
A certain sound level increases by an additional 20 dB. By how much does the intensity increase? The intensity increases by a factor of 200. The intensity increases by a factor of 100. The intensity increases by a factor of 20. The intensity increases by a factor of 2. The intensity does not increase.
The sound level changes by 10 dB , so Δ𝛽=20 dB . Decibels work on a multiplicative scale. Every 10 dB increase in sound level means a factor of 10 increase in the intensity. Since the sound level increases by 20 dB , there are two 10 dB increases, so the intensity is multiplied by a factor of 10 two times. This means that the intensity is multiplied by a factor of 100.
A block of mass 0.407 kg is hung from a vertical spring and allowed to reach equilibrium at rest. As a result, the spring is stretched by 0.610 m. Find the spring constant. =_____N/m The block is then pulled down an additional 0.316 m and released from rest. Assuming no damping, what is its period of oscillation? =_____s How high above the point of release does the block reach as it oscillates? =_____m
The spring constant is defined as 𝑘=𝐹/Δ𝑦 where 𝐹 denotes the stretching (or compressing) force and Δ𝑦 is the resulting displacement of the end of the spring. In the present case, the stretching force is the weight of the block, 𝐹=𝑚𝑔 where 𝑚 is the block's mass and 𝑔 denotes the acceleration due to gravity. Substitute this in the formula for the spring constant to obtain the result in terms of the given quantities. 𝑘=𝑚𝑔/Δ𝑦 Now substitute the numerical data to obtain the numerical value of the spring constant. 𝑘=(0.407 kg)(9.81 m/s2)/0.610 m=6.55 N/m The formula for the period of oscillation is 𝑇=2𝜋√𝑚/k Substitute the formula for the spring constant in terms of given quantities, which you obtained previously, and cancel the mass to obtain the period in terms of given quantities. 𝑇=2𝜋√Δ𝑦/𝑔 Substitute the given values. 𝑇=2𝜋√0.610 m/9.81 m/s^2 =1.57 s After its release from rest, the block oscillates about its equilibrium position in simple harmonic motion with an amplitude 𝐴, which equals the distance from which it was released below the equilibrium position. Therefore, it reaches the same distance, 𝐴, above the equilibrium position as it was released below the equilibrium position. In other words, it reaches a height of 2𝐴 above its release point. 2𝐴=2×(0.316 m)=0.632 m
The speed of a light wave in a certain transparent material is 0.639 times its speed in vacuum, which is 3.00×108 m/s . When yellow light with a frequency of 5.21×1014 Hz passes through this material, what is its wavelength 𝜆 in nanometers? 𝜆= ____nm
The wave speed 𝑣 is 𝑣=𝜙𝑐 where 𝜙 is the given factor and 𝑐 denotes the speed of light in vacuum, which is 3.00×10^8 m/s. For all periodic waves, the wave speed 𝑣, frequency 𝑓, and wavelength 𝜆 are related by 𝑓𝜆=𝑣 Solve for 𝜆 and substitute the expression for the wave speed in terms of the speed of light in vacuum. This gives the requested wavelength in terms of given quantities. 𝜆=𝑣/𝑓 = 𝜙𝑐/𝑓 To obtain the numerical answer, substitute the data and convert the result to nanometers as requested. 𝜆= 0.639(3.00×10^8 m/s) / (5.21×10^14 Hz) =3.68×10^−7 m =368 nm
A 971.9 kg car is moving at 21.9 m/s. Calculate the magnitude of its momentum.
We are given: mass and velocity. Equation: p=mv 971.9*21.9= 21284.61 kg*m/s
Would the contact force on the same hand be any different if the woman clapped her hands together, each with an initial speed of 5.25 m/s , if they come to rest in the same time interval of 2.25 ms ? O yes, because the initial momentum of the system will be different due to the second hand O yes, because the change in momentum is different O no, because the change in momentum for the first hand will be the same O It depends on the coeffcient of friction between the two hands. O no, because the second hand has zero momentum
no, because the change in momentum for the first hand will be the same
Identify the statements about springs as either true or false. As you compress a spring from equilibrium, the force the spring exerts on you increases.
true
Identify the statements about springs as either true or false. As you stretch a spring from equilibrium, the force the spring exerts on you increases.
true
A person sitting in a parked car hears an approaching ambulance siren at a frequency 𝑓1f1 . As it passes him and moves away, he hears a frequency 𝑓2f2 . What is the actual frequency 𝑓f of the source? 𝑓2<𝑓<𝑓1 𝑓=𝑓2+𝑓1 𝑓<𝑓2 𝑓>𝑓1 𝑓=𝑓2−𝑓1
𝑓2<𝑓<𝑓1
What is the momentum 𝑝hunter of the 85.0 kg hunter running at 7.55 m/s after missing the elephant?
𝑝hunter = 𝑚hunter*𝑣hunter =(85.0 kg)(7.55 m/s) =642 kg⋅m/s
A sinusoidal transverse wave travels along a long, stretched string. The amplitude of this wave is 0.0817 m, its frequency is 2.05 Hz, and its wavelength is 1.21 m. (a) What is the shortest transverse distance 𝑑 between a maximum and a minimum of the wave? 𝑑= ___m (b) How much time Δ𝑡 is required for 70.1 cycles of the wave to pass a stationary observer? Δ𝑡= ____s (c) Viewing the whole wave at any instant, how many cycles 𝑁 are there in a 37.1 m length of string? 𝑁= ____cycles
(a) The transverse distance 𝑑 between a maximum and a minimum of a sinusoidal transverse wave on a stretched string is twice the given amplitude 𝐴 of the wave. 𝑑=2𝐴=2(0.0817 m)=0.163 m (b) The period 𝑇 is the time required for a single cycle to pass a stationary observer. It is 𝑇=1𝑓 where 𝑓 denotes the given frequency. The requested time interval Δ𝑡 required for the given number 𝑛 of cycles to pass a stationary observer is thus Δ𝑡=𝑛𝑇=𝑛𝑓=70.1 cycles2.05 cycles/s=34.2 s (c) The length along the string of a single cycle is the given wavelength 𝜆. So the number 𝑁 of cycles in the given length 𝐿 of string is 𝑁=𝐿𝜆=37.1 m1.21 m/cycle=30.7 cycles
Select the higher harmonics of a string fixed at both ends that has a fundamental frequency of 80 Hz. 80 Hz 200 Hz 240 Hz 160 Hz 120 Hz
Higher harmonics are equal to integer multiples of the fundamental frequency. The only two frequencies that are integer multiples of 80 Hz are 160 Hz and 240 Hz.
If a source radiates sound uniformly in all directions and you triple your distance from the sound source, what happens to the sound intensity at your new position? The sound intensity increases to three times its original value. The sound intensity does not change. The sound intensity drops to 1/3 of its original value. The sound intensity drops to 1/9 of its original value. The sound intensity drops to 1/27 of its original value.
Intensity 𝐼 is power per unit area. The source radiates sound with power 𝑃o in all directions uniformly, so the area 𝐴 over which the sound energy is spread is the surface area of a sphere with a radius 𝑟 that is the distance between the source and the receiver. Assume the initial distance is 𝑟o . 𝐼=𝑃o/𝐴 =𝑃o/4𝜋𝑟2 Calculate the new intensity 𝐼new at triple the distance by using 3𝑟o as the radius. 𝐼new=𝑃o/4𝜋(3𝑟o)^2 =𝑃o/4𝜋(9𝑟^2o) =1/9(𝑃o/4𝜋𝑟^2o) =1/9𝐼o Intensity drops to 1/9 of its original value when the distance is tripled. This can be understood conceptually by recognizing that the area is proportional to the square of the distance, and that when the sound energy is being spread out over a greater area, the average intensity of the sound decreases.
A block on a frictionless surface is attached to a horizontal spring. The spring is stretched so the block is at rest at 𝑥=𝐴, then the spring is released. At what point in the resulting simple harmonic motion is the speed of the block at its maximum? 𝑥=𝐴 and 𝑥=−𝐴 𝑥=0 and 𝑥=𝐴 𝑥=0 and 𝑥=−𝐴 𝑥=0 𝑥=0, 𝑥=𝐴, and 𝑥=−𝐴
No conservative forces act on the system, so the total mechanical energy 𝐸 of the system is conserved. The total mechanical energy of the system is the sum of the spring potential energy 𝑈k and the kinetic energy 𝐾 of the block. 𝐸=𝑈k+𝐾 Express 𝑈k and 𝐾 in terms of the spring constant 𝑘, the distance from equilibrium 𝑥, and the block's mass 𝑚 and velocity 𝑣. 𝐸 = 1/2𝑘𝑥^2 + 1/2𝑚𝑣^2 The total energy of the system equals the initial potential energy of the block, 1/2𝑘𝐴^2. Therefore, when 𝑥=±𝐴, the system's energy is purely potential energy, so velocity equals 0 at those positions. Thus, 𝑥=0 is the only possible answer. Affirmatively, when 𝑥=0, the system's energy is entirely kinetic, and therefore its speed must be a maximum there.
Replacing an object attached to a spring with an object having 1/4 the original mass will change the frequency of oscillation of the system by a factor of 1/4 1/2 1 , or no change 2 4
Note that the spring constant remains the same, and only the mass of the oscillating object is changing. The angular frequency 𝜔 of an oscillator is given in terms of mass 𝑚 and spring constant 𝑘 by 𝜔 = √𝑘/𝑚 Take this to be the initial frequency and then determine the new frequency 𝜔new in terms of the new mass 𝑚new. 𝜔new = √𝑘/𝑚new Use the information given in the problem statement that 𝑚new=14𝑚 and plug that into the expression. 𝜔new = √4𝑘/𝑚 Now factor out the 4 from under the square root. 𝜔new = √2𝑘/𝑚 This expression is precisely two times the original frequency 𝜔. 𝜔new = 2𝜔
A mass attached to the end of a spring is set in motion. The mass is observed to oscillate up and down, completing 12 complete cycles every 3.00 s. What is the period 𝑇 of the oscillation? 𝑇= ____s What is the frequency 𝑓 of the oscillation? 𝑓= ____Hz
The period 𝑇 is the amount of time for the completion of one full cycle of the motion. In this problem, the mass completes 12 cycles in 3.00 seconds. Thus, the time for one cycle is the ratio of the given number of cycles to the given amount of time. 𝑇 = 3.00 s / 12 =0.250 s Therefore, the period of oscillation is 0.250 s. The frequency 𝑓 is the number of cycles per unit of time. Mathematically, the frequency 𝑓 is the inverse of the period. 𝑓= 1 / 𝑇 Substitute the given values, and evaluate. 𝑓=1/𝑇 =1/(3.00 s / 12) =4.00 Hz
A person slaps her leg with her hand, which results in her hand coming to rest in a time interval of 2.25 ms from an initial speed of 5.25 m/s . What is the magnitude of the average contact force exerted on the leg, assuming the total mass of the hand and the forearm to be 1.55 kg ?
To find the average force 𝐹avg, use its relationship to the change in momentum Δ𝑝 and the time interval Δ𝑡. Δ𝑝=𝐹avgΔ𝑡 Rearrange for the average force. 𝐹avg=Δ𝑝Δ𝑡 The change in momentum is given by Δ𝑝 = 𝑝f−𝑝i = 𝑚(𝑣f−𝑣i) where 𝑣i is the initial velocity, 𝑣f is the final velocity, and 𝑚 is the mass. Plugging this into the force equation gives 𝐹avg = 𝑚(𝑣𝑓−𝑣𝑖) / Δ𝑡 = −𝑚𝑣𝑖 / Δ𝑡 The last simplification follows from the fact that the final velocity is zero. The problem asks for the magnitude of the average force, so omit the negative sign from your calculation. Convert the time from milliseconds to seconds and then substitute the given values into the equation. 𝐹avg= (1.55 kg)(5.25 m/s) / 2.25×10−3 s =3620 N
What is the momentum of a garbage truck that is 13500 kg and is moving at 10.0 m/s? Momentum of truck: 135000 At what speed would a 9.15 kg trash can have the same momentum as the truck?
To find the velocity of the trash can such that the truck and the can have the same momentum, again use the definition of momentum. Rearranging for the velocity gives: 𝑣=𝑝𝑚 Set the momentum of the truck equal to the momentum of the can: 𝑝can = 𝑝trash = 1.35000 kg⋅m/s Finally, plugging in the numbers gives 𝑣can = 𝑝can/𝑚can =1.35000 / 9.15 =14800 m/s
Compare the elephant's momentum with the momentum 𝑝dartpdart of a 0.0395 kg0.0395 kg tranquilizer dart fired at a speed of 611 m/s611 m/s. O 𝑝elephant=𝑝dart O 𝑝elephant>𝑝dart O 𝑝elephant<𝑝dart O More information is needed to determine how the momentums compare.
calculate the momentum of the dart in the same way. 𝑝dart= 𝑚dart*𝑣dart =(0.0395 kg)(611 m/s) =24.1 kg⋅m/s The momentum of the elephant is much larger than that of the dart. 𝑝elephant>𝑝dart
A hungry 181 kg lion running northward at 79.1 km/hr attacks and holds onto a 37.1 kg Thomson's gazelle running eastward at 61.4 km/hr. Find the final speed of the lion-gazelle system immediately after the attack. final speed: _____m/s
𝑝⃗ i = 𝑝⃗ f where 𝑝i is the initial momentum of the system and 𝑝f is the final momentum of the system immediately before and after the collision, respectively. The momentum of the system is just the vector sum of the two objects colliding, and thus equation (1) can be written as 𝑝⃗ i,1+𝑝⃗ i,2 = 𝑝⃗ f,1+𝑝⃗ f,2(2) Finally, by using the definition of momentum, equation (2) becomes 𝑚1𝑣⃗ i,1+𝑚2𝑣⃗ i,2 = 𝑚1𝑣⃗ f,1+𝑚2𝑣⃗ f,2 Immediately after the collision, both animals move together as one object of mass 𝑚1+𝑚2 at a common, unknown final velocity, 𝑣f. 𝑚1𝑣⃗ i,1+𝑚2𝑣⃗ i,2 = (𝑚1+𝑚2)𝑣⃗ f you can break Equation (4) into components, 𝑥-component: 𝑚1𝑣i,1𝑥+𝑚2𝑣i,2𝑥 = (𝑚1+𝑚2)𝑣f,𝑥 𝑦-component: 𝑚1𝑣i,1𝑦+𝑚2𝑣i,2𝑦 = (𝑚1+𝑚2)𝑣f,𝑦 Changing the speeds of the lion and the gazelle to SI units, v1= 79.1 km/hr⋅1000 m/1 km⋅1 hr/3600 s =22.0 m/s v2= 61.4 km/hr⋅1000 m/1 km⋅1 hr/3600 s =17.1 m/s Using these values and the masses given in the problem, 𝑚1= 181 kg𝑣i,1𝑥 = 0𝑣i,1𝑦 = 22.0 m/s 𝑚2= 37.1 kg𝑣i,2𝑥 = 17.1 m/s𝑣i,2𝑦 = 0 Note that 𝑣i,1𝑥 and 𝑣i,2𝑦 are both zero because initially the lion is only moving northward along the positive 𝑦‑axis (no 𝑥‑component) and the gazelle is only moving eastward in the positive 𝑥‑direction (no 𝑦‑component). Substituting these values into the component equations, 𝑥‑component: (181 kg)⋅(0 m/s)+(37.1 kg)⋅(17.1 m/s) = (181 kg+37.1 kg) 633 kg⋅m/s / 218 kg 𝑣f,𝑥 =𝑣f,𝑥 = 2.90 m/s 𝑦-component: (181 kg)⋅(22.0 m/s)+(37.1 kg)⋅(0 m/s) = (181 kg+37.1 kg)𝑣f,𝑦 3980 kg⋅m/s218 kg=𝑣f,𝑦 =18.2 m/s Now that you have determined the 𝑥‑ and 𝑦‑ components of the final velocity, solve for the final speed by using the Pythagorean theorem. vf2=vf2,x + vf2,y |vf| = √vf2x + vf2y =√(2.90 m/s)^2+(18.2 m/s)^2 =18.5 m/s
A string has a mass of 13.3 g. The string is stretched with a force of 9.01 N, giving it a length of 1.93 m. Then, the string vibrates transversely at precisely the frequency that corresponds to its fourth normal mode; that is, at its fourth harmonic. What is the wavelength of the standing wave created in the string? wavelength:_____m What is the frequency of the standing wave? frequency:______Hz
𝜆𝑛 = 2𝐿/𝑛 𝑛=1, 2, 3, ... In the question, you are given 𝐿 and 𝑛. Substitute to obtain the numerical answer for the wavelength 𝜆4. 𝜆4=2(1.93 m)/4 =0.965 m The frequency of the standing wave in the 𝑛th normal mode, which is the 𝑛th harmonic 𝑓𝑛, can be found using the relation that is valid for all periodic waves, 𝑣=√𝐹/𝑚/𝐿 𝑓𝑛=𝑛/2𝐿√𝐹/𝑚/𝐿 𝑛=1, 2, 3, etc. For the value of the fourth harmonic, convert the mass to the base SI unit, kilograms, and substitute the data. 𝑓4= 4/2(1.93 m)√(9.01 N) / (0.0133 kg)/(1.93 m) =37.5 Hz
