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ICE Princess25 [194]
3 years ago
6

Two ice skaters, paula and ricardo, push off from each other. ricardo weighs more than paula. part a which skater, if either, ha

s the greater momentum after the push-off? which skater, if either, has the greater momentum after the push-off? paula has the greater momentum. ricardo has the greater momentum. both skaters have momentums of equal magnitude. submitmy answersgive up part b which skater, if either, has the greater speed after the push-off? which skater, if either, has the greater speed after the push-off? paula has the greater speed. ricardo has the greater speed. both skaters have equal speeds.
Physics
1 answer:
ahrayia [7]3 years ago
7 0

Apply the law of conservation of momentum for this situation. The law states that the momentum of a system is constant (in absence of external forces acting on it).

The 'system' in this case are the two skaters. There is no external force on the skaters. Suppose the skaters are initially standing still. The momentum in the system is 0. This value will need to remain constant, even after the mutual push (which is a set of forces from <em>inside</em> the system). So we know that

(total momentum before) = (total momentum after)

Indexing the masses and velocities by the first letter of the skaters' names:

0 = m_P\vec v_P+m_R\vec v_R\\m_P\vec v_P = m_R(-\vec v_R)

From the last row, you can see that the skaters will have momentum of same magnitude but opposite direction, after the push off. That answers the first question: neither will have a greater momentum (both will have one of same magnitude).

Since Ricardo is heavier, from the above equality it follows that

m_R>m_P\implies|\vec v_R|

In words, Paula has the greater speed, after the push-off.

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two object gravitationally attract with a force of 20N.if the distance between the two object centers is doubled, then what is t
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Projectile A is launched horizontally at a speed of 20. Meters per second from the top of a cliff and strikes a level surface be
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consider the motion of projectile A in vertical direction :

v₀ = initial velocity of projectile A in vertical direction = 0 m/s         (since the projectile was launched horizontally)

a = acceleration of the projectile = g = acceleration due to gravity = 9.8 m/s²

t = time of travel for projectile A = 3.0 seconds

Y = vertical displacement of projectile A = height of the cliff = h = ?

using the kinematics equation along the vertical direction as

Y = v₀ t + (0.5) a t²

h = (0) (3.0) + (0.5) (9.8) (3.0)²

h = 44.1 m

4 0
3 years ago
Sam is recklessly driving 60 mph in a 30 mph speed zone when he suddenly sees the police. he steps on the brakes and slows to 30
barxatty [35]
For this problem, we use the derived equations for rectilinear motion at constant acceleration. The equations used for this problem are:

a = (v - v₀)/t
2ax = v² - v₀²
where
a is the acceleration
x is the distance
v is the final velocity
v₀ is the initial velocity
t is the time

The solution is as follows;

a = (60mph - 30 mph)/(3 s * 1 h/3600 s)
a = 36,000 mph²

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5 0
2 years ago
1 2 3 4 5 6 7 8 9 10
7nadin3 [17]

Answer:

<h2>C. <u>0.55 m/s towards the right</u></h2>

Explanation:

Using the conservation of law of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.

Momentum = Mass (M) * Velocity(V)

BEFORE COLLISION

Momentum of 0.25kg body moving at 1.0m/s = 0.25*1 = 0.25kgm/s

Momentum of 0.15kg body moving at 0.0m/s(body at rest) = 0kgm/s

AFTER COLLISION

Momentum of 0.25kg body moving at x m/s = 0.25* x= 0.25x kgm/s

<u>x is the final velocity of the 0.25kg ball</u>

Momentum of 0.15kg body moving at 0.75m/s(body at rest) =

0.15 * 0.75kgm/s = 0.1125 kgm/s

Using the law of conservation of momentum;

0.25+0 = 0.25x + 0.1125

0.25x = 0.25-0.1125

0.25x = 0.1375

x = 0.1375/0.25

x = 0.55m/s

Since the 0.15 kg ball moves off to the right after collision, the 0.25 kg ball will move at <u>0.55 m/s towards the right</u>

<u></u>

4 0
2 years ago
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