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gulaghasi [49]
2 years ago
9

A 40kg girl and a 50kg boy are facing each other on a friction-free rollerskates. The girl pushes the boy, who moves away at a s

peed of 3 m/s. What is the girl's
speed?
Your answer
Physics
1 answer:
ki77a [65]2 years ago
5 0

Answer:

Explanation:

The Law of Momentum Conservation, just like the Law of Thermodynamics about energy, says that momentum is neither created nor destroyed but is conserved, meaning it has to go somewhere. If the girl pushes the boy and they are both on friction free skates, then the girl will also react to the push. Momentum Conservation says

(m_gv_g+m_bv_b)_b=(m_gv_g+m_bv_b)_a In words this says that the mass times the velocity of the girl plus the mass times the velocity of the boy before the push has to equal the mass times velocity of the girl plus the mass times velocity of the boy after the push. Mathematically,

(40.0*0+50.0*0)_b=(40.0v+50.0*-3)_a

The left side of this is equal to 0. On the right, I made the velocity of the boy negative. We could have made it positive and it wouldn't have mattered. The sign will only be important to the result because if the sign of the girl's velocity is the same as the boy's, she is moving in the same direction as he is; if it's different, she is moving in the opposite direction.

0 = 40.0v - 150.0 and

-40.0v = -150.0 so

v = 3.75 This means that when she pushes the boy one way, mometum is conserved and she moves in the opposite direction and at a greater velocity (because her mass is less). Physics is a wonderful thing, isn't it!?

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7. A child of mass m starts from rest and slides without friction from a height h along a curved waterslide (Fig. P5.46). She is
marissa [1.9K]

The mechanical energy of the girl will be conserved because the system is isolated and the initial potential energy will be equal to final kinetic energy.

<h3>What is the law of conservation of energy?</h3>

The law of conservation of energy states that energy can neither be created nor destroyed but can be transformed from one form to another.

The change in the potential energy of the  launched from a height into the pool without friction from the given height h is calculated by applying the following kinematic equation.

ΔP.E = ΔK.E

where;

  • ΔP.E is change in potential energy of the child
  • ΔK.E is change in the kinetic energy of the child

mghf - mghi = ¹/₂mv²  - ¹/₂mu²

where;

  • m is the mass of the girl
  • g is acceleration due to gravity
  • hi is the initial height of the girl
  • hf is the final height when she is launched into the pool
  • u is the initial velocity
  • v is the final velocity of the girl

Thus, for every closed or isolated system such as this case, mechanical energy is always conserved because the initial potential energy of the girl will be converted into her final kinetic energy.

Learn more about conservation of mechanical energy here: brainly.com/question/332163

#SPJ1

4 0
1 year ago
A 2190 kg car moving east at 10.5 m/s collides with a 3220 kg car moving east. The cars stick together and move east as a unit a
Bezzdna [24]

To solve this problem it is necessary to apply the concepts related to the conservation of the Momentum describing the inelastic collision of two bodies. By definition the collision between the two bodies is given as:

m_1v_1+m_2v_2 = (m_1+m_2)V_f

Where,

m_{1,2}= Mass of each object

v_{1,2}= Initial Velocity of Each object

V_f= Final Velocity

Our values are given as

m_1 = 2190Kg

v_1 =10.5m/s

m_2 = 3220kg

V_f = 4.74m/s

Replacing we have that

m_1v_1+m_2v_2 = (m_1+m_2)V_f

(2190)(10.5)+(3220)v_2 = (2190+3220)(4.74)

v_2 = 0.8224m/s

Therefore the the velocity of the 3220 kg car before the collision was 0.8224m/s

8 0
3 years ago
Cart 1 of mass m is traveling with speed 2vo in the +x-direction when it has an elastic collision with cart 2 of
iogann1982 [59]

Answer:

Explanation:

Momentum conservation

m2v_0+2mv_0=mv_1+2mv_2 \quad (1/m) \quad 4v_0=v_1+2v_2\\

Kinetic energy conservation

\displaystyle \frac{1}{2}m(2v_0)^2+\frac{1}{2}2mv_0^2=\frac{1}{2}mv_1^2+\frac{1}{2}2mv_2^2 \quad (1/m) \quad 6v_0^2=v_1^2+2v_2^2

Solve the system

6 0
3 years ago
A graph titled Position versus time for with horizontal axis time (seconds) and vertical axis position (meters). The line runs i
Lelu [443]

Answer:

The first one is 3 m/s

The second one is 2 m/s

Explanation:

8 0
2 years ago
Read 2 more answers
a stone with a mass of 2.40 kg is moving with velocity (6.60î − 2.40ĵ) m/s. find the net work (in j) on the stone if its velocit
ch4aika [34]

By the work energy theorem, the total work done on the stone is given by its change in kinetic energy,

W = \Delta K = \dfrac m2 ({v_2}^2 - {v_1}^2)

We have

\vec v_1 = (6.60\,\vec\imath - 2.40\,\vec\jmath)\dfrac{\rm m}{\rm s} \implies {v_1}^2 = \|\vec v_1\|^2 = 49.32 \dfrac{\rm m^2}{\rm s^2}

\vec v_2 = (8.00\,\vec\imath + 4.00\,\vec\jmath) \dfrac{\rm m}{\rm s} \implies {v_2}^2 = \|\vec v_2\|^2 = 80.0\dfrac{\mathrm m^2}{\mathrm s^2}

Then the total work is

W = \dfrac{2.40\,\rm kg}2 \left(80.0\dfrac{\rm m^2}{\rm s^2} - 49.32\dfrac{\rm m^2}{\rm s^2}\right)  \approx \boxed{36.8\,\rm J}

5 0
2 years ago
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