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Anni [7]
2 years ago
9

Mass m moves to the right with speed =v along a frictionless horizontal surface and crashes into an equal mass m initially at re

st. upon colliding, the two masses stick together and move with speed v to the right. notice that v and v denote different speeds. after the collision the magnitude of the momentum of the system is:
Physics
1 answer:
Amiraneli [1.4K]2 years ago
4 0

After the collision the magnitude of the momentum of the system is Mv

Given:

mass of 1st object = M

speed of 1st object = v

mass of 2nd object = M

speed of 2nd object = 0

To Find:

magnitude of the momentum after collision

Solution: Product of the mass of a particle and its velocity. Momentum is a vector quantity; i.e., it has both magnitude and direction. Isaac Newton's second law of motion states that the time rate of change of momentum is equal to the force acting on the particle.

Applying conservation of linear momentum

Mv + M(0) = 2MV

Mv = 2MV

V = v/2

So, after collision momentum is

p = 2MV = 2xMxv/2 = Mv

So, after collision momentum is Mv

Learn more about Momentum here:

brainly.com/question/1042017

#SPJ4

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The question is incomplete. It comes with a set of answer choices.


These are the answer choices:


Alex observes it as 10 m/s, and his friend observes it as less than 10 m/s.


Alex observes it as less than 10 m/s, and his friend observes it as 10 m/s.


Both Alex and his friend observe it as 10 m/s.


Both Alex and his friend observe it as less than 10 m/s.



Answer: Both Alex and his friend observe it as 10 m/s.


Justification:


1) The speed is relative to the frame of reference.


2) It is said that the both Alex and his friend are standing still.


3) Then, the speed they both see is the same, 10 m/s, respect the Earth (where they are standing still).


Of course, Alex is watching the ball moving away and his friend is seing it approaching, but it is not relevant for the question, as it deals with the speed which is only about magnitude, not direction.

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3 years ago
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A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle
xz_007 [3.2K]

Answer:

a) \Delta{t} = 5.39s

b) the motorcycle travels 155 m

Explanation:

Let t_2-t_1 = \Delta{t}, then consider the equation of motion for the motorcycle (accelerated) and for the car (non accelerated):

v_{m2}=v_0+a\Delta{t}\\x+d=(\frac{v_0+v_{m2}}{2} )\Delta{t}\\v_c = v_0 = \frac{x}{\Delta{t}}

where:

v_{m2} is the speed of the motorcycle at time 2

v_{c} is the velocity of the car (constant)

v_{0} is the velocity of the car and the motorcycle at time 1

d is the distance between the car and the motorcycle at time 1

x is the distance traveled by the car between time 1 and time 2

Solving the system of equations:

\left[\begin{array}{cc}car&motorcycle\\x=v_0\Delta{t}&x+d=(\frac{v_0+v_{m2}}{2}}) \Delta{t}\end{array}\right]

v_0\Delta{t}=\frac{v_0+v_{m2}}{2}\Delta{t}-d \\\frac{v_0+v_{m2}}{2}\Delta{t}-v_0\Delta{t}=d\\(v_0+v_{m2})\Delta{t}-2v_0\Delta{t}=2d\\(v_0+v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\(2v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\a\Delta{t}^2=2d\\\Delta{t}=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2*58}{4}}=\sqrt{29}=5.385s

For the second part, we need to calculate x+d, so you can use the equation of the car to calculate x:

x = v_0\Delta{t}= 18\sqrt{29}=96.933m\\then:\\x+d = 154.933

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Answer:

You never know if the medication could make you worse

Explanation:

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