what is momentum of a train that is 60,000 kg that is moving at velocity of 17m/s?
explain your answer
"45 meters north" is an example of
Answer:
Displacement
Explanation:
The quantity 45m north is a typical example of displacement.
Displacement is the distance traveled by a body in a specific direction. Displacement is a vector quantity with both magnitude and direction.
When we are specifying the displacement of a body, the direction must be indicated accurately. Therefore, the quantity given is displacementIf a person weighs 140 lb'on Earth, their mass in kilograms is
Answer:
70 kg
Explanation:
divide it by 2
Hope this helped!
Answer:
63.502932 Kilograms
Explanation:
A sprinter starts from rest and accelerated at a rate of 0.16 m/s over a distance of 50.0 meters. How fast is the athletes traveling at the end of the 50.0 meters?
Answer:
40m/s
Explanation:
v²=u²+2as
v²=0²+2(16)(50)
v²=160v=40m/s
One disadvantage to experimental research is that experimental conditions do not always reflect reality.
Please select the best answer from the choices provided
T
F
Answer:
It's true I took the test on Edge.
Explanation:
Answer:
True
Explanation:
Got it right on edg
An 8.00 kg mass moving east at 15.4 m/s on a frictionless horizontal surface collides with a 10.0 kg object that is initially at rest. After the collision, the 8.00 kg object moves south at 3.90 m/s. (a) What is the velocity of the 10.0 kg object after the collision
Answer:
9.3m/s
Explanation:
Based on the law of conservation of momentum
Sum of momentum before collision = sum of momentum after collision
m1u1 +m2u2 = m1v1+m2v2
m1 = 8kg
u1 = 15.4m/s
m2 = 10kg
u2 = 0m/s(at rest)
v1 = 3.9m/s
Required
v2.
Substitute
8(15.4)+10(0) = 8(3.9)+10v2
123.2=31.2+10v2
123.2-31.2 = 10v2
92 = 10v2
v2 = 92/10
v2 = 9.2m/s
Hence the velocity of the 10.0 kg object after the collision is 9.2m/s
A stone is dropped from the top of a high cliff with zero initial velocity. In which system is the net momentum zero as the stone falls freely
Answer:
A system that includes the stone and the earth.
Explanation:
If the system of being dropped from the height of the cliff consists of just the stone alone, then it means that its momentum will certainly undergo changes as it falls freely. However, If the system is now expanded to include not only the stone but also the Earth, then it implies that the momentum of the stone which is in the downward direction will be equal and opposite to the momentum of the Earth in the upwards direction towards the stone. Therefore, the momentum will cancel out and net momentum will be zero.
A system of stone and earth can result to a net zero momentum.
Conservation of linear momentum
The principle of conservation of linear momentum states that the sum of the initial momentum is equal to the sum of final momentum.
[tex]m_1u_1 + m_2 u_2 = m_1v_1 + m_2 v_2[/tex]
A system that consists a linear system of stone and earth can result to a net zero momentum.
Thus, a system of stone and earth can result to a net zero momentum.
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If there is "waste" energy, does the Law of Conservation of Energy still apply?
Explanation:
Yes, the law of conservation of energy still applies even if there is waste energy.
The waste energy are the transformation products of energy from one form to another.
According to the law of conservation of energy "energy is neither created nor destroyed by transformed from one form to another in a system".
But of then times, energy is lost as heat or sound within a system.
If we take into account these waste energy, we can see that energy is indeed conserved. The sum total of the energy generated and those produced will be the same if we factor in other forms in which the energy has been transformed into.A cylindrical tank of radius R, filled to the top with a liquid, has a small hole in the side, of radius r, at distance d below the surface. Find an expression for the volume flow rate through the hole. A 4.0-mm-diameter hole is 1.0 m below the surface of a 2.0-m-diameter tank of water. What is the rate, in mm/min, at which the water level will initially drop if the water is not replenished?
Answer:
a)[tex]V=\pi *r^2 * \sqrt{2gd}[/tex]
b)[tex]dh / dt = 0.2658 mm / min[/tex]
Explanation:
From the question we are told that
Diameter of hole [tex]d_h=4mm=>0.004m[/tex]
Depth of hole [tex]D=0mm=>0.001m[/tex]
Diameter of tank [tex]d_t=2mm=>0.002m[/tex]
Generally the equation for pressure is mathematically given as
[tex]Pressure P= \rho*g*d[/tex]
[tex]P= 1/2*\rho *v^2[/tex]
Where
[tex]v = \sqrt {2gd}[/tex]
[tex]V = Area*v[/tex]
[tex]V=\pi *r^2 * \sqrt{2gd}[/tex]
Generally the level at which the water level will initially drop if the water is not replenished is mathematically given by
[tex]dh / dt = (r/R)^2 *sqrt{2gd}\\dh / dt = (2/2000)^2 *sqrt(2*9.81*1) \\dh / dt = 4.429*10^-3 mm/s \\[/tex]
Therefore the level at which the water level will initially drop if the water is not replenished
[tex]dh / dt = 0.2658 mm / min[/tex]
The rate, in mm/min, at which the water level will initially drop will be 1.0625 mm/min.
Given data:
The diameter of hole is, d = 4.0 mm = 0.004 m.
The depth of hole is, h = 1.0 m.
The diameter of tank is, d' = 2.0 m.
The given problem is based on the flow rate, which is defined as the flow of liquid through a given section per unit time.
Let us first obtain the equation of pressure as,
[tex]P=\dfrac{1}{2} \times \rho \times v^{2}[/tex]
Here, v is the velocity of efflux and its value is,
[tex]v=\sqrt{2gh} \\\\v^{2}=2gh[/tex]
And the level at which the water level will initially drop if the water is not replenished is mathematically given by,
[tex]\dfrac{dH}{dt}=(r/R)^{2} \times v[/tex]
Here,
r is the radius of hole.
R is the radius of tank.
Solving as,
[tex]\dfrac{dH}{dt}=((d/2) /(d'/2))^{2} \times \sqrt{2gh} \\\\\dfrac{dH}{dt}=((0.004/2) /(2/2))^{2} \times \sqrt{2 \times 9.8 \times 1}\\\\\dfrac{dH}{dt}=1.77 \times 10^{-5} \;\rm m/s\\\\\dfrac{dH}{dt}=1.77 \times 10^{-5} \times 6 \times 10^{4} \;\rm mm/min\\\\\dfrac{dH}{dt}=1.0625 \;\rm mm/min[/tex]
Thus, we can conclude that the rate, in mm/min, at which the water level will initially drop will be 1.0625 mm/min.
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As waves crash into rock along the shoreline, particles of sand, shell, and other materials in the ocean water loosen tiny bits of sediment from the rock. As the waves recede, they carry the sediment away. In this scenario, which process represents weathering, and which process represents erosion?
Answer:
WEATHERING is represented by the scenario (As waves crash into rock along the shoreline, particles of sand, shell, and other materials in the ocean water loosen tiny bits of sediment from the rock).
Erosion is represented by the scenario (As the waves recede, they carry the sediment away).
Explanation:
A wave is a disturbance which travels through a medium and transfers energy from one point to another. When wind blows over a water body like the ocean, ocean waves are formed. As the generated energy from the wind is transported through the water by the waves, the can hit against rocks on the shores leading to its break down with time. WEATHERING occurs when tiny bit of sediments from rocks are loosened due to the impact of ocean waves.
Erosion can be described as the wearing away of the earth's surface due to the impact of wind, rainfall ( water) or waves. There are different types of erosion which is classified according it's cause of formation.
Wave erosion occurs when sediments such as sand, shell and other materials are carried to the shoreline by ocean waves. This erodes the shore over time as the sediments act like sandpapers.
A vertical piston-cylinder device contains a gas at a pressure of 100 kPa. The piston has a mass of 10 kg and a diameter for 14 cm. Pressure of the gas is to be increased by placing some weights on the piston. Determine the local atmospheric pressure and the mass of the weights that will doublethe pressure of the gas inside the cylinder.
Answer:
the local atmospheric pressure is 93.63 kPa
the mass of the weights is 156.9 kg
Explanation:
Given that;
Initial pressure of gas = 100 kPa
mass of piston = 10 kg and diameter = 14 cm = 0.14 m
g = 9.81 m/s²
Now,
P_gas = P_atm + P_piston
100 = P_atm + P_piston --------- let this equation 1
P_piston = M_piston × g / A = (10 × 9.81) / π/4×d²
P_piston = 98.1 / (π/4×( 0.14 )²)
P_piston = 98.1 / 0.01539 = 6374,269 Pa = 6.37 kPa
now, from equation 1
100 = P_atm + P_piston
we substitute
100 = P_atm + 6.37
P_atm = 100 - 6.37
P_atm = 93.63 kPa
Therefore, the local atmospheric pressure is 93.63 kPa
Now for pressure of the gas in the cylinder ⇒ 2×initial pressure
Pgas_2 = 2 × 100 = 200 kPa
Pgas_2 = P_atm + P_piston + P_weight
Pgas_2 = P_gas + P_weight
we substitute
200 kPa = 100 kPa + P_weight
P_weight = 200 kPa - 100 kPa
P_weight = 100 kPa = 100,000 Pa
Also;
P_weight = M×g / A
100,000 Pa = ( M × 9.81 ) / (π/4 × (0.14)²)
100,000 × 0.01539 = M × 9.81
1539 = M × 9.81
M = 1539 / 9.81
M = 156.9 kg
Therefore, the mass of the weights is 156.9 kg
The mass of 60 paper clips is 18.0 grams. What is the mass of one paper clip?
Answer:
3.333333333333333333333333333333333333333
Explanation:
3.3333333333333333333333333333333333
Which of the physical variables listed below will change when you change the area of the capacitor plates (while keeping the battery connected).
a. Capacitance
b. Charge on the plates
c. Voltage across the plates
d. Net electric field between the plates
e. Energy stored in the capacitor
Answer:
a. Capacitance
b. Charge on the plates
e. Energy stored in the capacitor
Explanation:
Let A be the area of the capacitor plate
The capacitance of a capacitor is given as;
[tex]C = \frac{Q}{V} = \frac{\epsilon _0 A}{d} \\\\[/tex]
where;
V is the potential difference between the plates
The charge on the plates is given as;
[tex]Q = \frac{V\epsilon _0 A}{d}[/tex]
The energy stored in the capacitor is given as;
[tex]E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} (\frac{\epsilon _0 A}{d} )V^2[/tex]
Thus, the physical variables listed that will change include;
a. Capacitance
b. Charge on the plates
e. Energy stored in the capacitor
if you watch football let me know who you think is going to win super bowl 55 and what do you think the score going to be Kansas city chiefs or tampa bay buccaneers
Answer:
I think the bucs are gonna win because Tom Brady is on their team and it's rigged
but maybe I'm just thinking negatively lol
take a picture of an object in your house, describe the
energy stores and transfers that happen with it. You can be as imaginative as you wish
with the object (choose something unusual), but the stores you identify and transfers
that happen must be real.
pls give me ideas of what to take a photo of for this I'm really stuck :(
A neutral metal bob is hanging on the bottom of a pendulum that is 15 cm long. A charged balloon is held near the metal bob and the pendulum is pulled up to a vertical angle of 20-deg. If the mass of the metal bob is 0.025kg, what is the charge on the balloon.
Answer:
Explanation:
See the figure attached
F is electrostatic force .
T cos20 = mg
T sin20 = F
Tan20 = F / mg
F = mg tan 20 = .025 x 9.8 tan20
= .09 N
Distance between bob and balloon
= 15 sin20 = 5.1 cm = .051 m
If q be the charge on balloon
F = 9 x 10⁹ x q² / .051²
= 3460 x 10⁹ q² = .09
q² = 26 x 10⁻⁶ x 10⁻⁹
q = 16.12 x 10⁻⁸ C .
what type of waves can only travel through a medium?
Answer:
Mechanical waves
Explanation:
Mechanical waves are the waves that can travel only through a medium. Mechanical waves are disturbance of matter and require medium to transfer the energy. There are three types of mechanical waves that include transverse wave, longitudinal wave and surface wave.
Some of the examples of mechanical waves are sound waves and seismic waves etcetera.
Hence, the correct answer is "Mechanical waves".
A physics student spends part of her day walking between classes or for recreation, during which time she expends energy at an average rate of 280 W. The remainder of the day she is sitting in class, studying or resting; during these activities, she expends energy at a rate of 100 W. If she expends a total of 1.1 x 10^7 J of energy in a 24 hour day, how much of the day did she spend walking
The time of the day she spent walking is equal to 3.70 hrs.
What is power?Power can be explained as the rate of doing work in unit time. The SI unit of measurement of power is J/s or Watt (W). Power can be described as a time based quantity. The mathematical expression for power can be represented as mentioned below.
Power = work/time
P = W/t
Given, the energy spends part of her day walking, Ew = 280 W
The energy is spent by sitting in the class, Es = 100 W
The total energy spends, Et = 1.1 × 10⁷J
[tex]E_w \times t + E_s(24\times 60\times 60-t)= 1.1 \times 10^7J[/tex]
[tex]280 \times t + 100(24\times 60\times 60-t)= 1.1 \times 10^7[/tex]
280 t + 0.86 × 10⁷ - 100 t = 1.1 × 10⁷
180 t = 0.24 × 10⁷
t = 0.24 × 10⁷/180 × 3600
t = 3.70 hr
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What Coulombs discovered almost 300
years ago
Answer:
ummm hehe this is my time to shine
Explanation:
MERICIA!!!!!!!!!!!!!!!!!!!!!!!
When the bowling ball has fallen halfway down the building (height = 20 m), it has a speed of 19.8 m/s.
How much potential energy does the bowling ball have?
How much kinetic energy does the bowling ball have?
How much total energy (potential + kinetic) does the bowling ball have?
Of the bowling ball’s total energy, is more in the form of potential or kinetic energy?
Answer:
I think the answer is 19.8 potential energy
Explanation:
NONE.
If there is "waste" energy, does the Law of Conservation of Energy still apply? please don't type something random if so i'll just report it.
Explanation:
Yes, the law of conservation of energy still applies even if there is waste energy.
The waste energy are the transformation products of energy from one form to another.
According to the law of conservation of energy "energy is neither created nor destroyed by transformed from one form to another in a system".
But of then times, energy is lost as heat or sound within a system.
If we take into account these waste energy, we can see that energy is indeed conserved. The sum total of the energy generated and those produced will be the same if we factor in other forms in which the energy has been transformed into.what is the direction of the third force that would cause the box to remain stationary on the ramp ?
An arrow pointing from the bottom of the ramp to the top, I assume it would be friction.
A truck travelling down the street suddenly brakes, applying a 14 N force over 3.5 seconds. What was the impulse over the given time.
Answer:
49 Ns
Explanation:
Given data
Force= 14N
time = 3.5seconds
Applying the expression for impulse
P= Ft
substitute
P=14*3.5
P=49 Ns
Hence the impulse is 49 Ns
Two children, each with a mass of 25.4 kg, are at fixed locations on a merry-go-round (a disk that spins about an axis perpendicular to the disk and through its center). One child is 0.78 m from the center of the merry-go-round, and the other is near the outer edge, 3.14 m from the center. With the merry-go-round rotating at a constant angular speed, the child near the edge is moving with translational speed of 11.5 m/s.
a. What is the angular speed of each child?
b. Through what angular distance does each child move in 5.0 s?
c. Through what distance in meters does each child move in 5.0 s?
d. What is the centripetal force experienced by each child as he or she holds on?
e. Which child has a more difficult time holding on?
Answer:
a) ω₁ = ω₂ = 3.7 rad/sec
b) Δθ₁ = Δθ₂ = 18.5 rad
c) d₁ = 14.5 m d₂ = 57.5 m
d) Fc1 = 273.9 N Fc2 = 1069.8 N
e) The boy near the outer edge.
Explanation:
a)
Since the merry-go-round is a rigid body, any point on it rotates at the same angular speed.However, linear speeds of points at different distances from the center, are different.Applying the definition of angular velocity, and the definition of angle, we can write the following relationship between the angular and linear speeds:[tex]v = \omega*r (1)[/tex]
Since we know the value of v for the child near the outer edge, and the value of r for this point, we can find the value of the angular speed, as follows:[tex]\omega = \frac{v_{out} }{r_{out} } = \frac{11.5m/s}{3.14m} = 3.7 rad/sec (2)[/tex]
As we have already said, ωout = ωin = 3.7 rad/secb)
Since the angular speed is the same for both childs, the angle rotated in the same time, will be the same for both also.Applying the definition of angular speed, as the rate of change of the angle rotated with respect to time, we can find the angle rotated (in radians) as follows:[tex]\Delta \theta = \omega * t = 3.7 rad/sec* 5.0 sec = 18.5 rad (3)[/tex]⇒ Δθ₁ = Δθ₂ = 18.5 rad.
c)
The linear distance traveled by each child, will be related with the linear speed of them.Knowing the value of the angular speed, and the distance from each boy to the center, we can apply (1) in order to get the linear speeds, as follows:[tex]v_{inn} = \omega * r_{inn} = 3.7 rad/sec * 0.78 m = 2.9 m/s (4)[/tex]
vout is a given of the problem ⇒ vout = 11. 5 m/s
Applying the definition of linear velocity, we can find the distance traveled by each child, as follows:[tex]d_{inn} = v_{inn} * t = 2.9m/s* 5.0 s = 14.5 m (5)[/tex]
[tex]d_{out} = v_{out} * t = 11.5 m/s* 5.0 s = 57.5 m (6)[/tex]
d)
The centripetal force experienced by each child is the force that keeps them on a circular movement, and can be written as follows:[tex]F_{c} = m*\frac{v^{2}}{r} (7)[/tex]
Replacing by the values of vin and rin, since m is a given, we can find Fcin (the force on the boy closer to the center) as follows:[tex]F_{cin} = m*\frac{v_{in}^{2}}{r_{in}} = 25.4 kg* \frac{(2.9m/s)^{2} }{0.78m} = 273.9 N (8)[/tex]
In the same way, we get Fcout (the force on the boy near the outer edge):[tex]F_{cout} = m*\frac{v_{out}^{2}}{r_{out}} = 25.4 kg* \frac{(11.5m/s)^{2} }{3.14m} = 1069.8 N (9)[/tex]
e)
The centripetal force that keeps the boys in a circular movement, is not a different type of force, and in this case, is given by the static friction force.The maximum friction force is given by the product of the coefficient of static friction times the normal force.Since the boys are not accelerated in the vertical direction, the normal force is equal and opposite to the force due to gravity, which is the weight.As both boys have the same mass, the normal force is also equal.This means that for both childs, the maximum possible static friction force, is the same, and given by the following expression:[tex]F_{frs} = \mu_{s} * m* g (10)[/tex]If this force is greater than the centripetal force, the boy will be able to hold on.So, as the centripetal force is greater for the boy close to the outer edge, he will have a more difficult time holding on.When researchers replicate a study, they are seeking to __________.
A.
prove that the hypothesis upon which the study was founded is untestable
B.
develop a new hypothesis
C.
change the study to provide new results
D.
support or reject the hypothesis upon which the study was founded
Please select the best answer from the choices provided
A
B
C
D
Answer:
D
Explanation:
right edge 2022
What kind of scattering (Rayleigh, Mie, or non-selective) would you expect to be most important when radiation of the specified wavelength encounters the following natural or anthropogenic particles?
Slides 16-31, Lecture 2 ought to help - slides 19, 24, and 31 are key.
Wavelength O2 molecules Smoke particles Cloud droplets Rain droplets
(size 10^-10 m) (size 0.3 (μm) (20 μm) (size 3 mm)
550 nm
11 μm
1600 nm
1 cm
Solution :
1. Rayleigh scattering takes place when the particle size is smaller than the wavelength (λ).
2. Mie scattering takes place when particle size is nearly equal to the wavelength (λ).
3. Non-selective scatter takes place when particle size in greater than the wavelength (λ).
We have the sizes of different particles :
[tex]$O_2 \rightarrow 10^{10} \ m $[/tex]
Smoke particles [tex]$\rightarrow 3 \times 10^{-7} \ m$[/tex]
Cloud droplets [tex]$\rightarrow 2 \times 10^{-5} \ m$[/tex]
Rain droplets [tex]$\rightarrow 3 \times 10^{-3} \ m$[/tex]
Wavelength [tex]$ O_2 $[/tex] Smoke particles Cloud droplets Rain droplets
[tex]$10^{-10} \ m$[/tex] [tex]$ 3 \times 10^{-7} \ m$[/tex] [tex]$ 2 \times 10^{-5} \ m$[/tex] [tex]$ 3 \times 10^{-3} \ m$[/tex]
[tex]$5500 \times 10^{-4} \ m$[/tex] Rayleigh Non-selective Non-selective Non-selective
[tex]$11 \times 10^{-6} \ m $[/tex] Rayleigh Rayleigh Non-selective Non-selective
[tex]$1600 \times 10^{-10} \ m $[/tex] Rayleigh Non-selective Non-selective Non-selective
[tex]$10^{-2} \ m $[/tex] Rayleigh Rayleigh Rayleigh Mie
A student is driving through a mountainous region where the road is at some times flat, at some times inclined upward, and at some time inclined downward. The student maintains a speed of 20 m/s on the roadway, but is required to make an emergency stop on the three sepearte occasions. On levels roadway, it takes 25 m to stop. On a downward-sloping roadway, it takes 40 m to stop. On an upward-sloping roadway, it takes 18 m to stop. Explain why the stopping distances are different. (Focus answer using work and energy, other concepts may be used as well but be sure work and energy are included.)
Answer:
Explanation:
It is frictional force of the ground that helps in bringing the vehicle to stop . In the process of stopping , negative work is done on the car by friction force to overcome its kinetic energy .
At levelled road , for stoppage
Kinetic energy of vehicle = Work done by frictional force . = friction force x displacement .
At upward slopping road , gravitational force acting downward also helps the vehicle to stop do friction has to do less work .
At upward inclined road , for stoppage
Kinetic energy of vehicle = Work done by frictional force + work done by gravitational force = (friction force + gravitational force ) x displacement .
Hence displacement is less .
At downward slopping road , friction has to do more work because friction has to do work against gravitational force acting downwards wards and kinetic energy of the vehicle also .
At downward inclined road , for stoppage
Kinetic energy of vehicle + work done by gravitational force = Work done by frictional force = friction force x displacement .
Hence displacement is more .
Hence displacement is more in the downward slopping.
What is Displacement?Displacement is defined as the change in position of an object. It is a vector quantity and has a direction and magnitude.
It is frictional force of the ground that helps in bringing the vehicle to stop . In the process of stopping , negative work is done on the car by friction force to overcome its kinetic energy .
At levelled road , for stoppage
Kinetic energy of vehicle = Work done by frictional force . = friction force x displacement .
At upward slopping road , gravitational force acting downward also helps the vehicle to stop do friction has to do less work .
At upward inclined road , for stoppage
Kinetic energy of vehicle = Work done by frictional force + work done by gravitational force = (friction force + gravitational force ) x displacement .
Hence displacement is less .
At downward slopping road , friction has to do more work because friction has to do work against gravitational force acting downwards wards and kinetic energy of the vehicle also .
At downward inclined road , for stoppage
Kinetic energy of vehicle + work done by gravitational force = Work done by frictional force = friction force x displacement .
Hence displacement is more in the downward slopping.
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is 0.8 kilograms bigger then 80 grams
Answer:
Yes
Explanation:
0.8 kilograms is equal to 800 grams
Answer:
Yes, 0.8 kilograms is greater than 80 grams
Explanation:
0.8 kilograms is equal to 800 grams and 80 grams is equal to 0.08 kilogrmas.
Sorry if I'm wrong, correct me.
g Incandescent bulbs generate visible light by heating up a thin metal filament to a very high temperature so that the thermal radiation from the filament becomes visible. One bulb filament has a surface area of 30 mm2 and emits 60 W when operating. If the bulb filament has an emissivity of 0.8, what is the operating temperature of the filament
Answer:
2577 K
Explanation:
Power radiated , P = σεAT⁴ where σ = Stefan-Boltzmann constant = 5.6704 × 10⁻⁸ W/m²K⁴, ε = emissivity of bulb filament = 0.8, A = surface area of bulb = 30 mm² = 30 × 10⁻⁶ m² and T = operating temperature of filament.
So, T = ⁴√(P/σεA)
Since P = 60 W, we substitute the vales of the variables into T. So,
T = ⁴√(P/σεA)
= ⁴√(60 W/(5.6704 × 10⁻⁸ W/m²K⁴ × 0.8 × 30 × 10⁻⁶ m²)
= ⁴√(60 W/(136.0896 × 10⁻¹⁴ W/K⁴)
= ⁴√(60 W/(13608.96 × 10⁻¹⁶ W/K⁴)
= ⁴√(0.00441 × 10¹⁶K⁴)
= 0.2577 × 10⁴ K
= 2577 K
Billiard ball A (0.35 kg) is struck such that it moves at 10 m/s toward a
second identical ball (Ball B). After the collision Ball A continues to move
in the same direction at 2 m/s. What is the magnitude of the velocity for
Ball B after the collision?
Before Collision:
10 m/s
A
After Collision:
2 m/s
O
Answer:
6m/s
Explanation:
Using the law of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the momentum after collision.
Using the expression
m1u1 + m2u2 = (m1+m2)v
m1 and m2 are the masses
u1 and u2 are the initial velocities
v is the final velocity after collision
Substitute the given values in the formula
0.35(10)+0.35(2) = (0.35+0.35)v
3.5+0.7 = 0.7v
4.2 = 0.7v
v = 4.2/0.7
v = 6m/s
Hence the magnitude of the velocity for Ball B after the collision is 6m/s
