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solniwko [45]
3 years ago
12

Ball X of mass 1.0 kg and ball Y of mass 0.5kg travel toward each other on a horizontal surface. Both balls travel with a consta

nt speed of 5 m/s until they collide. During the collision, ball Y exerts an average force with a magnitude of 40N for 1/6s on ball X . Which of the following best predicts ball momentum after the collision?
A. Ball Y will travel at a speed less than 5 m/s in the same direction of travel as before the collision.
B. Ball Y will travel at a speed less than 5 m/s in the opposite direction of travel as before the collision.
C. Ball will travel at a speed greater than 5 m/s in the same direction of travel as before the collision.
D. Ball Y motion cannot be predicted because the impulse on it is not known.
Physics
2 answers:
Alecsey [184]3 years ago
5 0

Answer:

C

Explanation:

Ball Y will travel at a speed greater than 5 m/s in the opposite direction of travel as before the collision.

bazaltina [42]3 years ago
4 0

Answer:

B. Ball Y will travel at a speed less than 5 m/s in the opposite direction of travel as before the collision.

Explanation:

Impulse created by ball Y  on ball X = 40 x 1/6 Ns

Ball X will also create impulse 40 / 6 on ball Y .

impulse = change in momentum .

impulse in Y = change in momentum in Y .

Initial momentum of Y  = .5 x 5 = 2.5

Let final velocity of Y after collision be v  in opposite direction .

change in momentum of Y = v - (-2.5 )

so,

v + 2.5 = 40 / 6 =  6.67

v =  4.17 m / s .

Option B is correct .

B. Ball Y will travel at a speed less than 5 m/s in the opposite direction of travel as before the collision.

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A high diver of mass 51.7 kg steps off a board 10.0 m above the water and falls vertical to the water, starting from rest. If he
zmey [24]

Answer:

851.33 N

Explanation:

Using newton;s equation of motion,

v² = u² + 2gh ......... Equation 1

Where v = final velocity, u = initial velocity, g = acceleration due to gravity, h = height

Given: u = 0 m/s(from rest), h = 10 m g = 9.8 m/s².

Substitute into equation 1

v² = 0² + 2(9.8)(10)

v² = 196

v = √196

v = 14 m/s.

Note: As the Diver touches the water,  u = 14 m/s and v = 0 m/s( stopped)

Using,

d = (v+u)t/2 .............. equation 2

Where d = distance moved by the diver in water before its motion stopped, t = time taken before it comes to rest

Given: v = 0 m/s, u = 14 m/s t = 2.10 s

Substitute into equation 2

d = (0+14)2.1/2

d = 14.7 m.

Finally

work done by the water to stop the diver = potential energy of the diver

F×d = mgh'................Equation 3

Where F = force of the diver in water, d = distance of the diver in the water, m = mass of the diver, g = acceleration due to gravity, h' = height of the diver from the point of fall to the point where he comes to rest

making F the subject of the equation,

F = mgh/d ............ Equation 4

Given: m = 51.7 kg, h = 10 m, g = 9.8 m/s², d = 14.7 m.

Substitute into equation 4

F = 51.7(10+14.7)(9.8)/14.7

F = 851.33 N

Hence the upward force the water exert on her = 851.33 N

8 0
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If you push a crate across a factory floor at constant speed in a constant direction, what is the magnitude of the force of fric
poizon [28]

Answer:

The magnitude of the force of friction equals the magnitude of my push

Explanation:

Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.

Let F = push and f = frictional force and f' = net force

F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0

So, F - f = 0

Thus, F = f

So, the magnitude of the force of friction equals the magnitude of my push.

3 0
2 years ago
The position-time graph of an object is found to be a straight line passing through the origin. What information about the motio
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-- The object either left or crossed the starting line exactly at time=0 .

-- The object has been traveling at constant speed for all time that
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Answer:

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