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aliina [53]
3 years ago
14

Two balls undergo a perfectly elastic head-on collision, with one ball initially at rest. If the incoming ball has a speed of 20

0 m/s . Part APart complete What is the final speed of the incoming ball if it is much more massive than the stationary ball?

Physics
2 answers:
Bingel [31]3 years ago
8 0

Answer:

Explanation:

Check the attachment for solution

andrew11 [14]3 years ago
8 0

Answer: The final speed of the incoming ball is approximately 200m/s

Explanation:

Using the law of conservation of momentum.

m1(u1) + m2u2 = m1v1 + m2v2

And also law of conservation of kinetic energy for elastic heads on collision we can derive the formula for elastic heads on collision which is given below:

For elastic heads on collision.

v1 = [( m1 - m2)/(m1+m2)] u1 ......1

v2 = [(2m1)/(m1+m2)]u1 ......2

Where,

m1 and m2 are the mass of the incoming and stationary ball respectively.

u1 and u2 are the initial speed of the incoming and stationary ball respectively.

v1 and v2 are the final speed of the incoming and stationary ball respectively.

a) to determine the final speed of the incoming ball using equation 1

v1 = [( m1 - m2)/(m1+m2)]u1

Since m1 >> m2

m1 - m2 ~= m1 and m1 +m2 ~= m1

So, equation 1 becomes

v1 ~= [m1/m1]u1

v1 ~= u1

Since u1 = 200m/s

v1 ~= 200m/s

Additional tips: using equation 2 we can derive the approximate final speed of the stationary ball following the same assumptions. If well solved v2 = 2u1 = 400m/s

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A physics student stands on a cliff overlooking a lake and decides to throw a softball to her friends in the water below. She th
aliina [53]

Answer:

58.5 m

Explanation:

First of all, we need to find the total time the ball takes to reach the water. This can be done by looking at the vertical motion only.

The initial vertical velocity of the ball is

u_y = u sin \theta

where

u = 21.5 m/s is the initial speed

\theta=33.5^{\circ} is the angle

Substituting,

u_y = (21.5) sin 33.5^{\circ} =11.9 m/s

The vertical position of the ball at time t is given by

y = h + u_y t + \frac{1}{2}gt^2

where

h = 13.5 m is the initial heigth

g = -9.8 m/s^2 is the acceleration of gravity (negative sign because it points downward)

The ball reaches the water when y = 0, so

0 = h + u_yt +\frac{1}{2}gt^2\\0 = 13.5 +11.9 t - 4.9t^2

Which gives two solutions: t = 3.27 s and t = -0.84 s. We discard the negative solution since it is meaningless.

The horizontal velocity of the ball is

u_y = u cos \theta = (21.5) cos 33.5^{\circ} =17.9 m/s

And since the motion along the horizontal direction is a uniform motion, we can find the horizontal distance travelled by the ball as follows:

d= u_x t = (17.9)(3.27)=58.5 m

3 0
2 years ago
A gang of robbers is escaping across city roofs at night. They come to the edge of one building and need to drop down to their g
REY [17]

Answer:

a) They will hit the ground with a speed of 19.6 m/s.

b) They are at a height of 20 m.

c) It is not a safe jump.

Explanation:

Hi there!

a) The equations of height and velocity in function of time of a free falling body are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the object at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering downward as negative direction).

v = velocity of the object at time t.

Using the equation of velocity, let's find the velocity at which they will hit the ground. The pebble is dropped (initial velocity = 0) and it takes 2 s to reach the ground:

v = v0 + g · t     (v0 = 0)

v = g · t

v = -9.8 m/s² · 2.0 s

v = -19.6 m/s

They will hit the ground with a speed of 19.6 m/s.

b)Now, we have to use the equation of height:

h = h0 + v0 · t + 1/2 · g · t²

If we place the origin of the frame of reference on the ground, we have to find the initial height (h0) knowing that at t = 2.0 s, h = 0 m

0 m = h0 - 1/2 · 9.8 m/s² · (2.0 s)²

h0 = 1/2 · 9.8 m/s² · (2.0 s)²

h0 = 20 m

They are at a height of 20 m.

c)According to a NASA paper (Issues on Human Acceleration Tolerance After Long-Duration Space Flights, figure 10), if you fall with a vertical velocity greater than 17 m/s it is unlikely that you will survive. So, it is not a safe jump.  

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An object of mass 6.00 kg falls with an aceleration of 8.00 m/s2. The magnitude of air resitance must be ____ N
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Using Newton's Second Law, we can find the air resistance. We know the net force is equal to mass times acceleration.F_{net} = m*a = (6.00kg)(8.00 \frac{m}{s^2}) = 48N 

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48N = (6.00kg)(9.81m/s^2) - F_{d} 

F_{d} = 10.86N


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