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Thepotemich [5.8K]
2 years ago
12

Two hockey players skating on essentially frictionless ice collide head-on. Madeleine, of mass 65.0 kg, is moving at 6.00 m/s to

the east just before the collision and at 3.00 m/s to the west just after the collision. Buffy, of mass 55.0 kg, is moving at 3.50 m/s to the east just after the collision. Find Buffy's velocity (magnitude and direction) just before the collision.
Physics
1 answer:
xeze [42]2 years ago
5 0

Explanation:

It is given that,

Mass of Madeleine, m_1=65\ kg

Initial speed of Madeleine, u_1=6\ m/s (due east)

Final speed of Madeleine, v_1=-3\ m/s (due west)

Mass of Buffy, m_2=55\ kg

Final speed of Buffy, v_2=3.5\ m/s (due east)

Let u_1 is the Buffy's velocity just before the collision. Using the conservation of linear momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

65\times 6+55\times u_2=65\times (-3)+55\times 3.5

u_2=-7.13\ m/s

So, the initial speed of the Buffy just before the collision is 7.13 m/s and it is moving due west. Hence, this is the required solution.

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A uniform 1.0-N meter stick is suspended horizontally by vertical strings attached at each end. A 2.0-N weight is suspended from
fgiga [73]

Answer:

3.5 N

Explanation:

Let the 0-cm end be the moment point. We know that for the system to be balanced, the total moment about this point must be 0. Let's calculate the moment at each point, in order from 0 to 100cm

- Tension of the string attached at the 0cm end is 0 as moment arm is 0

- 2 N weight suspended from the 10 cm position: 2*10 = 20 Ncm clockwise

- 2 N weight suspended from the 50 cm position: 2*50 = 100 Ncm clockwise

- 1 N stick weight at its center of mass, which is 50 cm position, since the stick is uniform: 1*50 = 50 Ncm clockwise

- 3 N weight suspended from the 60 cm position: 3*60 = 180 Ncm clockwise

- Tension T (N) of the string attached at the 100-cm end: T*100 = 100T Ncm counter-clockwise.

Total Clockwise moment = 20 + 100 + 50 + 180 = 350Ncm

Total counter-clockwise moment = 100T

For this to balance, 100 T = 350

so T = 350 / 100 = 3.5 N

4 0
1 year ago
A projectile of mass m is fired horizontally with an initial speed of v0​ from a height of h above a flat, desert surface. Negle
Grace [21]

Complete question is;

A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h and g:

(a) the work done by the force of gravity on the projectile,

(b) the change in kinetic energy of the projectile since it was fired, and

(c) the final kinetic energy of the projectile.

(d) Are any of the answers changed if the initial angle is changed?

Answer:

A) W = mgh

B) ΔKE = mgh

C) K2 = mgh + ½mv_o²

D) No they wouldn't change

Explanation:

We are expressing in terms of m, v0​, h, and g. They are;

m is mass

v0 is initial velocity

h is height of projectile fired

g is acceleration due to gravity

A) Now, the formula for workdone by force of gravity on projectile is;

W = F × h

Now, Force(F) can be expressed as mg since it is force of gravity.

Thus; W = mgh

Now, there is no mention of any angles of being fired because we are just told it was fired horizontally.

Therefore, even if the angle is changed, workdone will not change because the equation doesn't depend on the angle.

B) Change in kinetic energy is simply;

ΔKE = K2 - K1

Where K2 is final kinetic energy and K1 is initial kinetic energy.

However, from conservation of energy, we now that change in kinetic energy = change in potential energy.

Thus;

ΔKE = ΔPE

ΔPE = U2 - U1

U2 is final potential energy = mgh

U1 is initial potential energy = mg(0) = 0. 0 was used as h because at initial point no height had been covered.

Thus;

ΔKE = ΔPE = mgh

Again like a above, the change in kinetic energy will not change because the equation doesn't depend on the angle.

C) As seen in B above,

ΔKE = ΔPE

Thus;

½mv² - ½mv_o² = mgh

Where final kinetic energy, K2 = ½mv²

And initial kinetic energy = ½mv_o²

Thus;

K2 = mgh + ½mv_o²

Similar to a and B above, this will not change even if initial angle is changed

D) All of the answers wouldn't change because their equations don't depend on the angle.

5 0
2 years ago
What is the kinetic energy of a 100 kg object that is moving with a speed of 12.5m/s
Doss [256]

The kinetic energy of any moving object is

                           (1/2) (mass) (speed²) .

For the object you described, that's

                            (1/2) (100 kg) (12.5 m/s)²

                         =      (50 kg)  (156.25 m²/s²)

                         =              7,812.5 joules  
______________________________

Your attachment is way out of focus, and impossible to read.

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Which of the following sketches represents a possible configuration for this problem?
garri49 [273]
Where are the following sketches?
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OlgaM077 [116]
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