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lidiya [134]
3 years ago
8

Consider a system to be one train car moving toward another train car at rest. When the trains collide, the two cars stick toget

her. What is the total momentum of the system after the collision?
Before Collison
Car 1 Car 2
m1 =600kg m2 =400kg
v1 =4 m/s v2 =0 m/s
Choices:
800kg × m/s
1,600kg × m/s
2,400kg × m/s
4,000kg × m/s
Someone please help! The lady who tought me did a terrible job of explaining how to do this!
Physics
2 answers:
pickupchik [31]3 years ago
8 0

As per the question, the system consists of two cars.

The the masses of two cars are denoted as-  m_{1} \ and\ m_{2} \ respectively

The two cars undergoes collision with each other. Here collision must be inelastic in nature.

Let the velocities of each car is denoted as-  v_{1} \ and\ v_{2} \ respectively

It is given that masses of two cars and initial velocities-

m_{1} =600 kg          v_{1} =4m/s

m_{2} = 400kg         v_{2} =0 m/s

 

First we have to calculate the total momentum of the system before collision.

The momentum of a body is the product of its mass with velocity.

Momentum of car 1:

                           p_{1} =m_{1} v_{1}

                                  =600kg*4m/s

                                  =2400\ kg\ m\ s^-1

The momentum of car 2:

                             p_{2} =m_{2} v_{2}

                                   =400\ kg*0\ m/s

                                   =\ 0\ kg\ m\ s^-1

Now the total momentum of the system is :

                        p_{1} +p_{2}

                             =[2400+0]\ kg\ m\ s^-1

                             =2400\ kg\ m\ s^-1

Now we are asked to calculate the total momentum of the system after collision.

The basic condition in case of any type of collision is that it is in accordance with law of conservation of linear momentum.

Hence, the momentum will be conserved here.

It means  the total momentum before collision must be equal to the total momentum after collision.

Hence, the correct answer to the question will be 2400 kg m/s.



                 

3241004551 [841]3 years ago
6 0

Answer : 2400 kgm/s.

Explanation :

It is given that,

Mass of train 1, m_1=600\ kg

Mass of train 2, m_2=400\ kg

<em>When the trains collide, the two cars stick together. So, it is a perfectly inelastic collision. The momentum of the system will remain conserved.</em>

Momentum of train car 1, p_1=m_1v_1=2400\ Kgm/s

Momentum of train car 2, p_2=m_2v_2=0

So, total momentum of the system after the collision will be same as before the collision.

p_f=p_1+p_2

p_f=2400\ kgm/s

Hence, after the collision total momentum of the system would be 2400 kgm/s.

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A car accelerates uniformly from rest to speed 6.6 m/s in 6.5 s .Find the distance the car travel during this time .​
kirill [66]

Answer:

<em>The distance the car traveled is 21.45 m</em>

Explanation:

<u>Motion With Constant Acceleration </u>

It occurs when an object changes its velocity at the same rate thus the acceleration is constant.

The relation between the initial and final speeds is:

v_f=v_o+at\qquad\qquad [1]

Where:

a   = acceleration

vo = initial speed

vf  = final speed

t    = time

The distance traveled by the object is given by:

\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]

Solving [1] for a:

\displaystyle a=\frac{v_f-v_o}{t}

Substituting the given data vo=0, vf=6.6 m/s, t=6.5 s:

\displaystyle a=\frac{6.6-0}{6.5}

a = 1.015\ m/s^2

The distance is now calculated with [2]:

\displaystyle x=0*6.5+\frac{1.015*6.5^2}{2}

x = 21.45 m

The distance the car traveled is 21.45 m

6 0
3 years ago
You are traveling on an interstate highway at the posted speed limit of 70 mph when you see that the traffic in front of you has
vovikov84 [41]

Answer:8.75 s,

136.89 m

Explanation:

Given

Initial velocity=70 mph\approx 31.29 m/s

velocity after 5 s is 30 mph\approx 13.41 m/s

Therefore acceleration during these 5 s

a=\frac{v-u}{t}

a=\frac{13.41-31.29}{5}=-3.576 m/s^2

therefore time required to stop

v=u+at

here v=final velocity =0 m/s

initial velocity =31.29 m/s

0=31.29-3.576\times t

t=\frac{31.29}{3.576}=8.75 s

(b)total distance traveled before stoppage

v^2-u^2=2as

0^2-31.29^2=2\times (-3.576)\cdot s

s=136.89 m

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3 years ago
A tennis ball travelling at a speed of 46m/s with a mass of 58kg. Calculate the kinetic<br>energy​
Zanzabum

Answer:

its 1/2 the mass of the object times by its velocity ^ 2

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3 years ago
3. A football is kicked with a speed of 35 m/s at an angle of 40°.
jarptica [38.1K]

a) 22.5 m/s

The initial vertical velocity is given by:

u_y = u sin \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_y = (35)(sin 40)=22.5 m/s

b) 26.8 m/s

The initial horizontal velocity is given by:

u_x = u cos \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_x = (35)(cos 40)=26.8 m/s

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

v_y = u_y + at

where

v_y is the vertical velocity at time t

u_y = 22.5 m/s

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, v_y =0; substituting, we find the time t at which this happens:

0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

s=u_y t + \frac{1}{2}gt^2

where

s is the vertical displacement at time t

u_y = 22.5 m/s

g=-9.8 m/s^2

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

t=2(2.30 s)=4.60 s

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

v_x = 26.8 m/s

Therefore, the horizontal distance travelled during the whole motion is

d=v_x t = (26.8)(4.60)=123.3 m

So, the ball lands 123.3 m far from the initial point.

4 0
3 years ago
3. What do we call the ONLY part of the electromagnetic spectrum that we can
otez555 [7]

Answer:

Visible Light

wavelength = 4000 - 7000 Angstroms = 400 - 700 milli-microns

1 A unit =  10^-10 m

1 mμ = 10^-9 m

6 0
2 years ago
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